Buggy python submission (verdict: Runtime Error):
# for i, a in enumerate(iterable)
# q, mod = divmod(a, b)
# divmod(x, y) returns the tuple (x//y, x%y)
# manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
import sys
sys.setrecursionlimit(10**7)
from itertools import combinations, permutations # https://docs.python.org/ja/3/library/itertools.html
from math import factorial
def combinations_count(n, r):
    # faster than the following code
    # def combinations_count(n, r):
    #     return factorial(n) // (factorial(n - r) * factorial(r))
    if n - r < r: r = n - r
    if r == 0: return 1
    if r == 1: return n
    numerator = [n - r + k + 1 for k in range(r)]
    denominator = [k + 1 for k in range(r)]
    for p in range(2,r+1):
        pivot = denominator[p - 1]
        if pivot > 1:
            offset = (n - r) % p
            for k in range(p-1,r,p):
                numerator[k - offset] /= pivot
                denominator[k] /= pivot
    result = 1
    for k in range(r):
        if numerator[k] > 1:
            result *= int(numerator[k])
    return result
def combination_with_repetition_count(n, r):
    return combinations_count(n + r - 1, r)
from collections import deque, Counter # https://docs.python.org/ja/3/library/collections.html#collections.deque
from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
from functools import reduce
from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
def gcds(numbers):
    return reduce(gcd, numbers)
def lcm(x, y):
    return (x * y) // gcd(x, y)
def lcms(numbers):
    return reduce(lcm, numbers, 1)

# first create factorial_list!
# fac_list = mod_factorial_list(n)
INF = 10 ** 18
MOD = 10 ** 6 + 3
def modpow(a, n, p = MOD):
    # if   n == 0     : return 1
    # elif n == 1     : return a % p
    # elif n % 2 == 0 : return modpow(a, n//2, p)**2 % p
    # else            : return modpow(a, n//2, p)**2 * a % p
    lambda a, n: pow(a, n, p = MOD)
def modinv(a, p = MOD):
    # evaluate reciprocal using Fermat's little theorem:
    # a**(p-1) is identical to 1 (mod p) when a and p is coprime
    return modpow(a, p-2, p)
def modinv_list(n, p = MOD):
    if n <= 1:
        return [0,1][:n+1]
    else:
        inv_t = [0,1]
        for i in range(2, n+1):
            inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
        return inv_t
def modfactorial_list(n, p = MOD):
    if n == 0:
        return [1]
    else:
        l = [0] * (n+1)
        tmp = 1
        for i in range(1, n+1):
            tmp = tmp * i % p
            l[i] = tmp
        return l
def modcomb(n, k, fac_list = [], p = MOD):
    # fac_list = modfactorial_list(100)
    # print(modcomb(100, 5, modfactorial_list(100)))
    from math import factorial
    if n < 0 or k < 0 or n < k: return 0
    if n == 0 or k == 0: return 1
    if len(fac_list) <= n:
        a = factorial(n) % p
        b = factorial(k) % p
        c = factorial(n-k) % p
    else:
        a = fac_list[n]
        b = fac_list[k]
        c = fac_list[n-k]
    return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
def modadd(a, b, p = MOD):
    return (a + b) % MOD
def modsub(a, b, p = MOD):
    return (a - b) % p
def modmul(a, b, p = MOD):
    return ((a % p) * (b % p)) % p
def moddiv(a, b, p = MOD):
    return modmul(a, modpow(b, p-2, p))

# set the inputs
    # open(0).read() is a convenient method:
    # ex) n, m, *x = map(int, open(0).read().split())
    #     min(x[::2]) - max(x[1::2])
    # ex2) *x, = map(int, open(0).read().split())
    #     don't forget to add comma after *x if only one variable is used
    # R = lambda: list(map(int, input2().split()))
    # old_R = lambda: list(map(int, input().split()))
    # ex1) n, k = R()
    # ex2) v = list(R())
    # ex3) bc = [list(R()) for _ in range(m)]

# q = 2
# xdn = [[7, 2, 4], [12345, 67890, 2019]]

input2 = sys.stdin.readline
r = sys.stdin.readline #single: int(r()), line: map(int, r().split())
R = lambda: list(map(int, input2().split()))
q = int(input())
xdn = [R() for _ in range(q)]

# calculate
fac_list = modfactorial_list(MOD)
for x, d, n in xdn:
    if d == 0:
        print(modpow(x, n))
    else:
        x = moddiv(x, d)
        if x + n - 1 >= MOD:
            print(0)
        else:
            print(modmul(moddiv(fac_list[x+n-1], fac_list[x-1]),modpow(d, n)))
# output
# ex1) print(*l) => when l = [2, 5], printed 2 5 6


Fixed submission:
# for i, a in enumerate(iterable)
# q, mod = divmod(a, b)
# divmod(x, y) returns the tuple (x//y, x%y)
# manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
import sys
sys.setrecursionlimit(10**7)
from itertools import combinations, permutations # https://docs.python.org/ja/3/library/itertools.html
from math import factorial
def combinations_count(n, r):
    # faster than the following code
    # def combinations_count(n, r):
    #     return factorial(n) // (factorial(n - r) * factorial(r))
    if n - r < r: r = n - r
    if r == 0: return 1
    if r == 1: return n
    numerator = [n - r + k + 1 for k in range(r)]
    denominator = [k + 1 for k in range(r)]
    for p in range(2,r+1):
        pivot = denominator[p - 1]
        if pivot > 1:
            offset = (n - r) % p
            for k in range(p-1,r,p):
                numerator[k - offset] /= pivot
                denominator[k] /= pivot
    result = 1
    for k in range(r):
        if numerator[k] > 1:
            result *= int(numerator[k])
    return result
def combination_with_repetition_count(n, r):
    return combinations_count(n + r - 1, r)
from collections import deque, Counter # https://docs.python.org/ja/3/library/collections.html#collections.deque
from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
from functools import reduce
from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
def gcds(numbers):
    return reduce(gcd, numbers)
def lcm(x, y):
    return (x * y) // gcd(x, y)
def lcms(numbers):
    return reduce(lcm, numbers, 1)

# first create factorial_list!
# fac_list = mod_factorial_list(n)
INF = 10 ** 18
MOD = 10 ** 6 + 3
def modpow(a, n, p = MOD):
    if   n == 0     : return 1
    elif n == 1     : return a % p
    elif n % 2 == 0 : return modpow(a, n//2, p)**2 % p
    else            : return modpow(a, n//2, p)**2 * a % p
def modinv(a, p = MOD):
    # evaluate reciprocal using Fermat's little theorem:
    # a**(p-1) is identical to 1 (mod p) when a and p is coprime
    return modpow(a, p-2, p)
def modinv_list(n, p = MOD):
    if n <= 1:
        return [0,1][:n+1]
    else:
        inv_t = [0,1]
        for i in range(2, n+1):
            inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
        return inv_t
def modfactorial_list(n, p = MOD):
    if n == 0:
        return [1]
    else:
        l = [0] * (n+1)
        tmp = 1
        for i in range(1, n+1):
            tmp = tmp * i % p
            l[i] = tmp
        return l
def modcomb(n, k, fac_list = [], p = MOD):
    # fac_list = modfactorial_list(100)
    # print(modcomb(100, 5, modfactorial_list(100)))
    from math import factorial
    if n < 0 or k < 0 or n < k: return 0
    if n == 0 or k == 0: return 1
    if len(fac_list) <= n:
        a = factorial(n) % p
        b = factorial(k) % p
        c = factorial(n-k) % p
    else:
        a = fac_list[n]
        b = fac_list[k]
        c = fac_list[n-k]
    return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
def modadd(a, b, p = MOD):
    return (a + b) % MOD
def modsub(a, b, p = MOD):
    return (a - b) % p
def modmul(a, b, p = MOD):
    return ((a % p) * (b % p)) % p
def moddiv(a, b, p = MOD):
    return modmul(a, modpow(b, p-2, p))

# set the inputs
    # open(0).read() is a convenient method:
    # ex) n, m, *x = map(int, open(0).read().split())
    #     min(x[::2]) - max(x[1::2])
    # ex2) *x, = map(int, open(0).read().split())
    #     don't forget to add comma after *x if only one variable is used
    # R = lambda: list(map(int, input2().split()))
    # old_R = lambda: list(map(int, input().split()))
    # ex1) n, k = R()
    # ex2) v = list(R())
    # ex3) bc = [list(R()) for _ in range(m)]

# q = 2
# xdn = [[7, 2, 4], [12345, 67890, 2019]]

input2 = sys.stdin.readline
R = lambda: list(map(int, input2().split()))
q = int(input())
xdn = [R() for _ in range(q)]

# calculate
fac_list = modfactorial_list(MOD)
for x, d, n in xdn:
    if d == 0:
        print(modpow(x, n))
    else:
        x = moddiv(x, d)
        if x + n - 1 >= MOD:
            print(0)
        else:
            print(modmul(moddiv(fac_list[x+n-1], fac_list[x-1]),modpow(d, n)))
# output
# ex1) print(*l) => when l = [2, 5], printed 2 5 6

Buggy python submission (verdict: Time Limit Exceeded):
# for i, a in enumerate(iterable)
# q, mod = divmod(a, b)
# divmod(x, y) returns the tuple (x//y, x%y)
# manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
import sys
sys.setrecursionlimit(10**7)
from itertools import combinations, permutations # https://docs.python.org/ja/3/library/itertools.html
from math import factorial
def combinations_count(n, r):
    # faster than the following code
    # def combinations_count(n, r):
    #     return factorial(n) // (factorial(n - r) * factorial(r))
    if n - r < r: r = n - r
    if r == 0: return 1
    if r == 1: return n
    numerator = [n - r + k + 1 for k in range(r)]
    denominator = [k + 1 for k in range(r)]
    for p in range(2,r+1):
        pivot = denominator[p - 1]
        if pivot > 1:
            offset = (n - r) % p
            for k in range(p-1,r,p):
                numerator[k - offset] /= pivot
                denominator[k] /= pivot
    result = 1
    for k in range(r):
        if numerator[k] > 1:
            result *= int(numerator[k])
    return result
def combination_with_repetition_count(n, r):
    return combinations_count(n + r - 1, r)
from collections import deque, Counter # https://docs.python.org/ja/3/library/collections.html#collections.deque
from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
from functools import reduce
from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
def gcds(numbers):
    return reduce(gcd, numbers)
def lcm(x, y):
    return (x * y) // gcd(x, y)
def lcms(numbers):
    return reduce(lcm, numbers, 1)

# first create factorial_list!
# fac_list = mod_factorial_list(n)
INF = 10 ** 18
MOD = 10 ** 6 + 3
def modpow(a, n, p = MOD):
    if   n == 0     : return 1
    elif n == 1     : return a % p
    elif n % 2 == 0 : return modpow(a, n//2, p)**2 % p
    else            : return modpow(a, n//2, p)**2 * a % p
def modinv(a, p = MOD):
    # evaluate reciprocal using Fermat's little theorem:
    # a**(p-1) is identical to 1 (mod p) when a and p is coprime
    return modpow(a, p-2, p)
def modinv_list(n, p = MOD):
    if n <= 1:
        return [0,1][:n+1]
    else:
        inv_t = [0,1]
        for i in range(2, n+1):
            inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
        return inv_t
def modfactorial_list(n, p = MOD):
    if n == 0:
        return [1]
    else:
        l = [1]
        tmp = 1
        for i in range(1, n+1):
            tmp = tmp * i % p
            l.append(tmp)
        return l
def modcomb(n, k, fac_list = [], p = MOD):
    # fac_list = modfactorial_list(100)
    # print(modcomb(100, 5, modfactorial_list(100)))
    from math import factorial
    if n < 0 or k < 0 or n < k: return 0
    if n == 0 or k == 0: return 1
    if len(fac_list) <= n:
        a = factorial(n) % p
        b = factorial(k) % p
        c = factorial(n-k) % p
    else:
        a = fac_list[n]
        b = fac_list[k]
        c = fac_list[n-k]
    return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
def modadd(a, b, p = MOD):
    return (a + b) % MOD
def modsub(a, b, p = MOD):
    return (a - b) % p
def modmul(a, b, p = MOD):
    return ((a % p) * (b % p)) % p
def moddiv(a, b, p = MOD):
    return modmul(a, modpow(b, p-2, p))

# set the inputs
    # open(0).read() is a convenient method:
    # ex) n, m, *x = map(int, open(0).read().split())
    #     min(x[::2]) - max(x[1::2])
    # ex2) *x, = map(int, open(0).read().split())
    #     don't forget to add comma after *x if only one variable is used
    # R = lambda: list(map(int, input2().split()))
    # old_R = lambda: list(map(int, input().split()))
    # ex1) n, k = R()
    # ex2) v = list(R())
    # ex3) bc = [list(R()) for _ in range(m)]

# q = 2
# xdn = [[7, 2, 4], [12345, 67890, 2019]]

input2 = sys.stdin.readline
R = lambda: list(map(int, input2().split()))
q = int(input())
xdn = [R() for _ in range(q)]

# calculate
fac_list = modfactorial_list(MOD)
for (x, d, n) in xdn:
    if d == 0:
        print(modpow(x, n))
    else:
        x = moddiv(x, d)
        if x + n - 1 >= MOD:
            print(0)
        else:
            print(modmul(moddiv(fac_list[x+n-1], fac_list[x-1]),modpow(d, n)))
# output
# ex1) print(*l) => when l = [2, 5], printed 2 5 6


Fixed submission:
# for i, a in enumerate(iterable)
# q, mod = divmod(a, b)
# divmod(x, y) returns the tuple (x//y, x%y)
# manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
import sys
sys.setrecursionlimit(10**7)
from itertools import combinations, permutations # https://docs.python.org/ja/3/library/itertools.html
from math import factorial
def combinations_count(n, r):
    # faster than the following code
    # def combinations_count(n, r):
    #     return factorial(n) // (factorial(n - r) * factorial(r))
    if n - r < r: r = n - r
    if r == 0: return 1
    if r == 1: return n
    numerator = [n - r + k + 1 for k in range(r)]
    denominator = [k + 1 for k in range(r)]
    for p in range(2,r+1):
        pivot = denominator[p - 1]
        if pivot > 1:
            offset = (n - r) % p
            for k in range(p-1,r,p):
                numerator[k - offset] /= pivot
                denominator[k] /= pivot
    result = 1
    for k in range(r):
        if numerator[k] > 1:
            result *= int(numerator[k])
    return result
def combination_with_repetition_count(n, r):
    return combinations_count(n + r - 1, r)
from collections import deque, Counter # https://docs.python.org/ja/3/library/collections.html#collections.deque
from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
from functools import reduce
from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
def gcds(numbers):
    return reduce(gcd, numbers)
def lcm(x, y):
    return (x * y) // gcd(x, y)
def lcms(numbers):
    return reduce(lcm, numbers, 1)

# first create factorial_list!
# fac_list = mod_factorial_list(n)
INF = 10 ** 18
MOD = 10 ** 6 + 3
def modpow(a, n, p = MOD):
    if   n == 0     : return 1
    elif n == 1     : return a % p
    elif n % 2 == 0 : return modpow(a, n//2, p)**2 % p
    else            : return modpow(a, n//2, p)**2 * a % p
def modinv(a, p = MOD):
    # evaluate reciprocal using Fermat's little theorem:
    # a**(p-1) is identical to 1 (mod p) when a and p is coprime
    return modpow(a, p-2, p)
def modinv_list(n, p = MOD):
    if n <= 1:
        return [0,1][:n+1]
    else:
        inv_t = [0,1]
        for i in range(2, n+1):
            inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
        return inv_t
def modfactorial_list(n, p = MOD):
    if n == 0:
        return [1]
    else:
        l = [0] * (n+1)
        tmp = 1
        for i in range(1, n+1):
            tmp = tmp * i % p
            l[i] = tmp
        return l
def modcomb(n, k, fac_list = [], p = MOD):
    # fac_list = modfactorial_list(100)
    # print(modcomb(100, 5, modfactorial_list(100)))
    from math import factorial
    if n < 0 or k < 0 or n < k: return 0
    if n == 0 or k == 0: return 1
    if len(fac_list) <= n:
        a = factorial(n) % p
        b = factorial(k) % p
        c = factorial(n-k) % p
    else:
        a = fac_list[n]
        b = fac_list[k]
        c = fac_list[n-k]
    return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
def modadd(a, b, p = MOD):
    return (a + b) % MOD
def modsub(a, b, p = MOD):
    return (a - b) % p
def modmul(a, b, p = MOD):
    return ((a % p) * (b % p)) % p
def moddiv(a, b, p = MOD):
    return modmul(a, modpow(b, p-2, p))

# set the inputs
    # open(0).read() is a convenient method:
    # ex) n, m, *x = map(int, open(0).read().split())
    #     min(x[::2]) - max(x[1::2])
    # ex2) *x, = map(int, open(0).read().split())
    #     don't forget to add comma after *x if only one variable is used
    # R = lambda: list(map(int, input2().split()))
    # old_R = lambda: list(map(int, input().split()))
    # ex1) n, k = R()
    # ex2) v = list(R())
    # ex3) bc = [list(R()) for _ in range(m)]

# q = 2
# xdn = [[7, 2, 4], [12345, 67890, 2019]]

input2 = sys.stdin.readline
R = lambda: list(map(int, input2().split()))
q = int(input())
xdn = [R() for _ in range(q)]

# calculate
fac_list = modfactorial_list(MOD)
for x, d, n in xdn:
    if d == 0:
        print(modpow(x, n))
    else:
        x = moddiv(x, d)
        if x + n - 1 >= MOD:
            print(0)
        else:
            print(modmul(moddiv(fac_list[x+n-1], fac_list[x-1]),modpow(d, n)))
# output
# ex1) print(*l) => when l = [2, 5], printed 2 5 6

Buggy python submission (verdict: Runtime Error):
# for i, a in enumerate(iterable)
# q, mod = divmod(a, b)
# divmod(x, y) returns the tuple (x//y, x%y)
# manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
import sys
sys.setrecursionlimit(10**7)
from itertools import combinations, permutations # https://docs.python.org/ja/3/library/itertools.html
from math import factorial
def combinations_count(n, r):
    # faster than the following code
    # def combinations_count(n, r):
    #     return factorial(n) // (factorial(n - r) * factorial(r))
    if n - r < r: r = n - r
    if r == 0: return 1
    if r == 1: return n
    numerator = [n - r + k + 1 for k in range(r)]
    denominator = [k + 1 for k in range(r)]
    for p in range(2,r+1):
        pivot = denominator[p - 1]
        if pivot > 1:
            offset = (n - r) % p
            for k in range(p-1,r,p):
                numerator[k - offset] /= pivot
                denominator[k] /= pivot
    result = 1
    for k in range(r):
        if numerator[k] > 1:
            result *= int(numerator[k])
    return result
def combination_with_repetition_count(n, r):
    return combinations_count(n + r - 1, r)
from collections import deque, Counter # https://docs.python.org/ja/3/library/collections.html#collections.deque
from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
from functools import reduce
from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
def gcds(numbers):
    return reduce(gcd, numbers)
def lcm(x, y):
    return (x * y) // gcd(x, y)
def lcms(numbers):
    return reduce(lcm, numbers, 1)

# first create factorial_list!
# fac_list = mod_factorial_list(n)
INF = 10 ** 18
MOD = 10 ** 6 + 3
def modpow(a, n, p = MOD):
    if   n == 0     : return 1
    elif n == 1     : return a % p
    elif n % 2 == 0 : return modpow(a, n//2, p)**2 % p
    else            : return modpow(a, n//2, p)**2 * a % p
def modinv(a, p = MOD):
    # evaluate reciprocal using Fermat's little theorem:
    # a**(p-1) is identical to 1 (mod p) when a and p is coprime
    return modpow(a, p-2, p)
def modinv_list(n, p = MOD):
    if n <= 1:
        return [0,1][:n+1]
    else:
        inv_t = [0,1]
        for i in range(2, n+1):
            inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
        return inv_t
def modfactorial_list(n, p = MOD):
    if n == 0:
        return [1]
    else:
        l = [1]
        tmp = 1
        for i in range(1, n+1):
            tmp = tmp * i % p
            l.append(tmp)
        return l
def modcomb(n, k, fac_list = [], p = MOD):
    # fac_list = modfactorial_list(100)
    # print(modcomb(100, 5, modfactorial_list(100)))
    from math import factorial
    if n < 0 or k < 0 or n < k: return 0
    if n == 0 or k == 0: return 1
    if len(fac_list) <= n:
        a = factorial(n) % p
        b = factorial(k) % p
        c = factorial(n-k) % p
    else:
        a = fac_list[n]
        b = fac_list[k]
        c = fac_list[n-k]
    return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
def modadd(a, b, p = MOD):
    return (a + b) % MOD
def modsub(a, b, p = MOD):
    return (a - b) % p
def modmul(a, b, p = MOD):
    return ((a % p) * (b % p)) % p
def moddiv(a, b, p = MOD):
    return modmul(a, modpow(b, p-2, p))

# set the inputs
    # open(0).read() is a convenient method:
    # ex) n, m, *x = map(int, open(0).read().split())
    #     min(x[::2]) - max(x[1::2])
    # ex2) *x, = map(int, open(0).read().split())
    #     don't forget to add comma after *x if only one variable is used
    # R = lambda: list(map(int, input2().split()))
    # old_R = lambda: list(map(int, input().split()))
    # ex1) n, k = R()
    # ex2) v = list(R())
    # ex3) bc = [list(R()) for _ in range(m)]

# q = 2
# xdn = [[7, 2, 4], [12345, 67890, 2019]]

input2 = sys.stdin.readline
R = lambda: list(map(int, input2().split()))
q = int(input())
xdn = [R() for _ in range(q)]

# calculate
fac_list = modfactorial_list(MOD*2)
for (x, d, n) in xdn:
    x = moddiv(x, d)
    print(modmul(moddiv(fac_list[x+n-1], fac_list[x-1]),modpow(d, n)))
# output
# ex1) print(*l) => when l = [2, 5], printed 2 5 6


Fixed submission:
# for i, a in enumerate(iterable)
# q, mod = divmod(a, b)
# divmod(x, y) returns the tuple (x//y, x%y)
# manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
import sys
sys.setrecursionlimit(10**7)
from itertools import combinations, permutations # https://docs.python.org/ja/3/library/itertools.html
from math import factorial
def combinations_count(n, r):
    # faster than the following code
    # def combinations_count(n, r):
    #     return factorial(n) // (factorial(n - r) * factorial(r))
    if n - r < r: r = n - r
    if r == 0: return 1
    if r == 1: return n
    numerator = [n - r + k + 1 for k in range(r)]
    denominator = [k + 1 for k in range(r)]
    for p in range(2,r+1):
        pivot = denominator[p - 1]
        if pivot > 1:
            offset = (n - r) % p
            for k in range(p-1,r,p):
                numerator[k - offset] /= pivot
                denominator[k] /= pivot
    result = 1
    for k in range(r):
        if numerator[k] > 1:
            result *= int(numerator[k])
    return result
def combination_with_repetition_count(n, r):
    return combinations_count(n + r - 1, r)
from collections import deque, Counter # https://docs.python.org/ja/3/library/collections.html#collections.deque
from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
from functools import reduce
from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
def gcds(numbers):
    return reduce(gcd, numbers)
def lcm(x, y):
    return (x * y) // gcd(x, y)
def lcms(numbers):
    return reduce(lcm, numbers, 1)

# first create factorial_list!
# fac_list = mod_factorial_list(n)
INF = 10 ** 18
MOD = 10 ** 6 + 3
def modpow(a, n, p = MOD):
    if   n == 0     : return 1
    elif n == 1     : return a % p
    elif n % 2 == 0 : return modpow(a, n//2, p)**2 % p
    else            : return modpow(a, n//2, p)**2 * a % p
def modinv(a, p = MOD):
    # evaluate reciprocal using Fermat's little theorem:
    # a**(p-1) is identical to 1 (mod p) when a and p is coprime
    return modpow(a, p-2, p)
def modinv_list(n, p = MOD):
    if n <= 1:
        return [0,1][:n+1]
    else:
        inv_t = [0,1]
        for i in range(2, n+1):
            inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
        return inv_t
def modfactorial_list(n, p = MOD):
    if n == 0:
        return [1]
    else:
        l = [0] * (n+1)
        tmp = 1
        for i in range(1, n+1):
            tmp = tmp * i % p
            l[i] = tmp
        return l
def modcomb(n, k, fac_list = [], p = MOD):
    # fac_list = modfactorial_list(100)
    # print(modcomb(100, 5, modfactorial_list(100)))
    from math import factorial
    if n < 0 or k < 0 or n < k: return 0
    if n == 0 or k == 0: return 1
    if len(fac_list) <= n:
        a = factorial(n) % p
        b = factorial(k) % p
        c = factorial(n-k) % p
    else:
        a = fac_list[n]
        b = fac_list[k]
        c = fac_list[n-k]
    return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
def modadd(a, b, p = MOD):
    return (a + b) % MOD
def modsub(a, b, p = MOD):
    return (a - b) % p
def modmul(a, b, p = MOD):
    return ((a % p) * (b % p)) % p
def moddiv(a, b, p = MOD):
    return modmul(a, modpow(b, p-2, p))

# set the inputs
    # open(0).read() is a convenient method:
    # ex) n, m, *x = map(int, open(0).read().split())
    #     min(x[::2]) - max(x[1::2])
    # ex2) *x, = map(int, open(0).read().split())
    #     don't forget to add comma after *x if only one variable is used
    # R = lambda: list(map(int, input2().split()))
    # old_R = lambda: list(map(int, input().split()))
    # ex1) n, k = R()
    # ex2) v = list(R())
    # ex3) bc = [list(R()) for _ in range(m)]

# q = 2
# xdn = [[7, 2, 4], [12345, 67890, 2019]]

input2 = sys.stdin.readline
R = lambda: list(map(int, input2().split()))
q = int(input())
xdn = [R() for _ in range(q)]

# calculate
fac_list = modfactorial_list(MOD)
for x, d, n in xdn:
    if d == 0:
        print(modpow(x, n))
    else:
        x = moddiv(x, d)
        if x + n - 1 >= MOD:
            print(0)
        else:
            print(modmul(moddiv(fac_list[x+n-1], fac_list[x-1]),modpow(d, n)))
# output
# ex1) print(*l) => when l = [2, 5], printed 2 5 6

Buggy python submission (verdict: Runtime Error):
# for i, a in enumerate(iterable)
# q, mod = divmod(a, b)
# divmod(x, y) returns the tuple (x//y, x%y)
# manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
import sys
sys.setrecursionlimit(10**7)
from itertools import combinations, permutations # https://docs.python.org/ja/3/library/itertools.html
from math import factorial
def combinations_count(n, r):
    # faster than the following code
    # def combinations_count(n, r):
    #     return factorial(n) // (factorial(n - r) * factorial(r))
    if n - r < r: r = n - r
    if r == 0: return 1
    if r == 1: return n
    numerator = [n - r + k + 1 for k in range(r)]
    denominator = [k + 1 for k in range(r)]
    for p in range(2,r+1):
        pivot = denominator[p - 1]
        if pivot > 1:
            offset = (n - r) % p
            for k in range(p-1,r,p):
                numerator[k - offset] /= pivot
                denominator[k] /= pivot
    result = 1
    for k in range(r):
        if numerator[k] > 1:
            result *= int(numerator[k])
    return result
def combination_with_repetition_count(n, r):
    return combinations_count(n + r - 1, r)
from collections import deque, Counter # https://docs.python.org/ja/3/library/collections.html#collections.deque
from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
from functools import reduce
from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
def gcds(numbers):
    return reduce(gcd, numbers)
def lcm(x, y):
    return (x * y) // gcd(x, y)
def lcms(numbers):
    return reduce(lcm, numbers, 1)

# first create factorial_list!
# fac_list = mod_factorial_list(n)
INF = 10 ** 18
MOD = 10 ** 6 + 3
def modpow(a, n, p = MOD):
    if   n == 0     : return 1
    elif n == 1     : return a % p
    elif n % 2 == 0 : return modpow(a, n//2, p)**2 % p
    else            : return modpow(a, n//2, p)**2 * a % p
def modinv(a, p = MOD):
    # evaluate reciprocal using Fermat's little theorem:
    # a**(p-1) is identical to 1 (mod p) when a and p is coprime
    return modpow(a, p-2, p)
def modinv_list(n, p = MOD):
    if n <= 1:
        return [0,1][:n+1]
    else:
        inv_t = [0,1]
        for i in range(2, n+1):
            inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
        return inv_t
def modfactorial_list(n, p = MOD):
    if n == 0:
        return [1]
    else:
        l = [1]
        tmp = 1
        for i in range(1, n+1):
            tmp = tmp * i % p
            l.append(tmp)
        return l
def modcomb(n, k, fac_list = [], p = MOD):
    # fac_list = modfactorial_list(100)
    # print(modcomb(100, 5, modfactorial_list(100)))
    from math import factorial
    if n < 0 or k < 0 or n < k: return 0
    if n == 0 or k == 0: return 1
    if len(fac_list) <= n:
        a = factorial(n) % p
        b = factorial(k) % p
        c = factorial(n-k) % p
    else:
        a = fac_list[n]
        b = fac_list[k]
        c = fac_list[n-k]
    return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
def modadd(a, b, p = MOD):
    return (a + b) % MOD
def modsub(a, b, p = MOD):
    return (a - b) % p
def modmul(a, b, p = MOD):
    return ((a % p) * (b % p)) % p
def moddiv(a, b, p = MOD):
    return modmul(a, modpow(b, p-2, p))

# set the inputs
    # open(0).read() is a convenient method:
    # ex) n, m, *x = map(int, open(0).read().split())
    #     min(x[::2]) - max(x[1::2])
    # ex2) *x, = map(int, open(0).read().split())
    #     don't forget to add comma after *x if only one variable is used
    # R = lambda: list(map(int, input2().split()))
    # old_R = lambda: list(map(int, input().split()))
    # ex1) n, k = R()
    # ex2) v = list(R())
    # ex3) bc = [list(R()) for _ in range(m)]

# q = 2
# xdn = [[7, 2, 4], [12345, 67890, 2019]]

input2 = sys.stdin.readline
R = lambda: list(map(int, input().split()))
q = int(input())
xdn = [R() for _ in range(q)]

# calculate
fac_list = modfactorial_list(MOD)
for (x, d, n) in xdn:
    x = moddiv(x, d)
    print(modmul(moddiv(fac_list[x+n-1], fac_list[x-1]),modpow(d, n)))
# output
# ex1) print(*l) => when l = [2, 5], printed 2 5 6


Fixed submission:
# for i, a in enumerate(iterable)
# q, mod = divmod(a, b)
# divmod(x, y) returns the tuple (x//y, x%y)
# manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
import sys
sys.setrecursionlimit(10**7)
from itertools import combinations, permutations # https://docs.python.org/ja/3/library/itertools.html
from math import factorial
def combinations_count(n, r):
    # faster than the following code
    # def combinations_count(n, r):
    #     return factorial(n) // (factorial(n - r) * factorial(r))
    if n - r < r: r = n - r
    if r == 0: return 1
    if r == 1: return n
    numerator = [n - r + k + 1 for k in range(r)]
    denominator = [k + 1 for k in range(r)]
    for p in range(2,r+1):
        pivot = denominator[p - 1]
        if pivot > 1:
            offset = (n - r) % p
            for k in range(p-1,r,p):
                numerator[k - offset] /= pivot
                denominator[k] /= pivot
    result = 1
    for k in range(r):
        if numerator[k] > 1:
            result *= int(numerator[k])
    return result
def combination_with_repetition_count(n, r):
    return combinations_count(n + r - 1, r)
from collections import deque, Counter # https://docs.python.org/ja/3/library/collections.html#collections.deque
from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
from functools import reduce
from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
def gcds(numbers):
    return reduce(gcd, numbers)
def lcm(x, y):
    return (x * y) // gcd(x, y)
def lcms(numbers):
    return reduce(lcm, numbers, 1)

# first create factorial_list!
# fac_list = mod_factorial_list(n)
INF = 10 ** 18
MOD = 10 ** 6 + 3
def modpow(a, n, p = MOD):
    if   n == 0     : return 1
    elif n == 1     : return a % p
    elif n % 2 == 0 : return modpow(a, n//2, p)**2 % p
    else            : return modpow(a, n//2, p)**2 * a % p
def modinv(a, p = MOD):
    # evaluate reciprocal using Fermat's little theorem:
    # a**(p-1) is identical to 1 (mod p) when a and p is coprime
    return modpow(a, p-2, p)
def modinv_list(n, p = MOD):
    if n <= 1:
        return [0,1][:n+1]
    else:
        inv_t = [0,1]
        for i in range(2, n+1):
            inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
        return inv_t
def modfactorial_list(n, p = MOD):
    if n == 0:
        return [1]
    else:
        l = [0] * (n+1)
        tmp = 1
        for i in range(1, n+1):
            tmp = tmp * i % p
            l[i] = tmp
        return l
def modcomb(n, k, fac_list = [], p = MOD):
    # fac_list = modfactorial_list(100)
    # print(modcomb(100, 5, modfactorial_list(100)))
    from math import factorial
    if n < 0 or k < 0 or n < k: return 0
    if n == 0 or k == 0: return 1
    if len(fac_list) <= n:
        a = factorial(n) % p
        b = factorial(k) % p
        c = factorial(n-k) % p
    else:
        a = fac_list[n]
        b = fac_list[k]
        c = fac_list[n-k]
    return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
def modadd(a, b, p = MOD):
    return (a + b) % MOD
def modsub(a, b, p = MOD):
    return (a - b) % p
def modmul(a, b, p = MOD):
    return ((a % p) * (b % p)) % p
def moddiv(a, b, p = MOD):
    return modmul(a, modpow(b, p-2, p))

# set the inputs
    # open(0).read() is a convenient method:
    # ex) n, m, *x = map(int, open(0).read().split())
    #     min(x[::2]) - max(x[1::2])
    # ex2) *x, = map(int, open(0).read().split())
    #     don't forget to add comma after *x if only one variable is used
    # R = lambda: list(map(int, input2().split()))
    # old_R = lambda: list(map(int, input().split()))
    # ex1) n, k = R()
    # ex2) v = list(R())
    # ex3) bc = [list(R()) for _ in range(m)]

# q = 2
# xdn = [[7, 2, 4], [12345, 67890, 2019]]

input2 = sys.stdin.readline
R = lambda: list(map(int, input2().split()))
q = int(input())
xdn = [R() for _ in range(q)]

# calculate
fac_list = modfactorial_list(MOD)
for x, d, n in xdn:
    if d == 0:
        print(modpow(x, n))
    else:
        x = moddiv(x, d)
        if x + n - 1 >= MOD:
            print(0)
        else:
            print(modmul(moddiv(fac_list[x+n-1], fac_list[x-1]),modpow(d, n)))
# output
# ex1) print(*l) => when l = [2, 5], printed 2 5 6

Buggy python submission (verdict: Runtime Error):
# for i, a in enumerate(iterable)
# q, mod = divmod(a, b)
# divmod(x, y) returns the tuple (x//y, x%y)
# manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
import sys
sys.setrecursionlimit(10**7)
from itertools import combinations, permutations # https://docs.python.org/ja/3/library/itertools.html
from math import factorial
def combinations_count(n, r):
    # faster than the following code
    # def combinations_count(n, r):
    #     return factorial(n) // (factorial(n - r) * factorial(r))
    if n - r < r: r = n - r
    if r == 0: return 1
    if r == 1: return n
    numerator = [n - r + k + 1 for k in range(r)]
    denominator = [k + 1 for k in range(r)]
    for p in range(2,r+1):
        pivot = denominator[p - 1]
        if pivot > 1:
            offset = (n - r) % p
            for k in range(p-1,r,p):
                numerator[k - offset] /= pivot
                denominator[k] /= pivot
    result = 1
    for k in range(r):
        if numerator[k] > 1:
            result *= int(numerator[k])
    return result
def combination_with_repetition_count(n, r):
    return combinations_count(n + r - 1, r)
from collections import deque, Counter # https://docs.python.org/ja/3/library/collections.html#collections.deque
from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
from functools import reduce
from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
def gcds(numbers):
    return reduce(gcd, numbers)
def lcm(x, y):
    return (x * y) // gcd(x, y)
def lcms(numbers):
    return reduce(lcm, numbers, 1)

# first create factorial_list!
# fac_list = mod_factorial_list(n)
INF = 10 ** 18
MOD = 10 ** 6 + 3
def modpow(a, n, p = MOD):
    if   n == 0     : return 1
    elif n == 1     : return a % p
    elif n % 2 == 0 : return modpow(a, n//2, p)**2 % p
    else            : return modpow(a, n//2, p)**2 * a % p
def modinv(a, p = MOD):
    # evaluate reciprocal using Fermat's little theorem:
    # a**(p-1) is identical to 1 (mod p) when a and p is coprime
    return modpow(a, p-2, p)
def modinv_list(n, p = MOD):
    if n <= 1:
        return [0,1][:n+1]
    else:
        inv_t = [0,1]
        for i in range(2, n+1):
            inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
        return inv_t
def modfactorial_list(n, p = MOD):
    if n == 0:
        return [1]
    else:
        l = [1]
        tmp = 1
        for i in range(1, n+1):
            tmp = tmp * i % p
            l.append(tmp)
        return l
def modcomb(n, k, fac_list = [], p = MOD):
    # fac_list = modfactorial_list(100)
    # print(modcomb(100, 5, modfactorial_list(100)))
    from math import factorial
    if n < 0 or k < 0 or n < k: return 0
    if n == 0 or k == 0: return 1
    if len(fac_list) <= n:
        a = factorial(n) % p
        b = factorial(k) % p
        c = factorial(n-k) % p
    else:
        a = fac_list[n]
        b = fac_list[k]
        c = fac_list[n-k]
    return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
def modadd(a, b, p = MOD):
    return (a + b) % MOD
def modsub(a, b, p = MOD):
    return (a - b) % p
def modmul(a, b, p = MOD):
    return ((a % p) * (b % p)) % p
def moddiv(a, b, p = MOD):
    return modmul(a, modpow(b, p-2, p))

# set the inputs
    # open(0).read() is a convenient method:
    # ex) n, m, *x = map(int, open(0).read().split())
    #     min(x[::2]) - max(x[1::2])
    # ex2) *x, = map(int, open(0).read().split())
    #     don't forget to add comma after *x if only one variable is used
    # R = lambda: list(map(int, input2().split()))
    # old_R = lambda: list(map(int, input().split()))
    # ex1) n, k = R()
    # ex2) v = list(R())
    # ex3) bc = [list(R()) for _ in range(m)]

# q = 2
# xdn = [[7, 2, 4], [12345, 67890, 2019]]

input2 = sys.stdin.readline
R = lambda: list(map(int, input2().split()))
q = int(input())
xdn = [R() for _ in range(q)]

# calculate
fac_list = modfactorial_list(MOD)
for (x, d, n) in xdn:
    x = moddiv(x, d)
    print(modmul(moddiv(fac_list[x+n-1], fac_list[x-1]),modpow(d, n)))
# output
# ex1) print(*l) => when l = [2, 5], printed 2 5 6


Fixed submission:
# for i, a in enumerate(iterable)
# q, mod = divmod(a, b)
# divmod(x, y) returns the tuple (x//y, x%y)
# manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
import sys
sys.setrecursionlimit(10**7)
from itertools import combinations, permutations # https://docs.python.org/ja/3/library/itertools.html
from math import factorial
def combinations_count(n, r):
    # faster than the following code
    # def combinations_count(n, r):
    #     return factorial(n) // (factorial(n - r) * factorial(r))
    if n - r < r: r = n - r
    if r == 0: return 1
    if r == 1: return n
    numerator = [n - r + k + 1 for k in range(r)]
    denominator = [k + 1 for k in range(r)]
    for p in range(2,r+1):
        pivot = denominator[p - 1]
        if pivot > 1:
            offset = (n - r) % p
            for k in range(p-1,r,p):
                numerator[k - offset] /= pivot
                denominator[k] /= pivot
    result = 1
    for k in range(r):
        if numerator[k] > 1:
            result *= int(numerator[k])
    return result
def combination_with_repetition_count(n, r):
    return combinations_count(n + r - 1, r)
from collections import deque, Counter # https://docs.python.org/ja/3/library/collections.html#collections.deque
from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
from functools import reduce
from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
def gcds(numbers):
    return reduce(gcd, numbers)
def lcm(x, y):
    return (x * y) // gcd(x, y)
def lcms(numbers):
    return reduce(lcm, numbers, 1)

# first create factorial_list!
# fac_list = mod_factorial_list(n)
INF = 10 ** 18
MOD = 10 ** 6 + 3
def modpow(a, n, p = MOD):
    if   n == 0     : return 1
    elif n == 1     : return a % p
    elif n % 2 == 0 : return modpow(a, n//2, p)**2 % p
    else            : return modpow(a, n//2, p)**2 * a % p
def modinv(a, p = MOD):
    # evaluate reciprocal using Fermat's little theorem:
    # a**(p-1) is identical to 1 (mod p) when a and p is coprime
    return modpow(a, p-2, p)
def modinv_list(n, p = MOD):
    if n <= 1:
        return [0,1][:n+1]
    else:
        inv_t = [0,1]
        for i in range(2, n+1):
            inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
        return inv_t
def modfactorial_list(n, p = MOD):
    if n == 0:
        return [1]
    else:
        l = [0] * (n+1)
        tmp = 1
        for i in range(1, n+1):
            tmp = tmp * i % p
            l[i] = tmp
        return l
def modcomb(n, k, fac_list = [], p = MOD):
    # fac_list = modfactorial_list(100)
    # print(modcomb(100, 5, modfactorial_list(100)))
    from math import factorial
    if n < 0 or k < 0 or n < k: return 0
    if n == 0 or k == 0: return 1
    if len(fac_list) <= n:
        a = factorial(n) % p
        b = factorial(k) % p
        c = factorial(n-k) % p
    else:
        a = fac_list[n]
        b = fac_list[k]
        c = fac_list[n-k]
    return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
def modadd(a, b, p = MOD):
    return (a + b) % MOD
def modsub(a, b, p = MOD):
    return (a - b) % p
def modmul(a, b, p = MOD):
    return ((a % p) * (b % p)) % p
def moddiv(a, b, p = MOD):
    return modmul(a, modpow(b, p-2, p))

# set the inputs
    # open(0).read() is a convenient method:
    # ex) n, m, *x = map(int, open(0).read().split())
    #     min(x[::2]) - max(x[1::2])
    # ex2) *x, = map(int, open(0).read().split())
    #     don't forget to add comma after *x if only one variable is used
    # R = lambda: list(map(int, input2().split()))
    # old_R = lambda: list(map(int, input().split()))
    # ex1) n, k = R()
    # ex2) v = list(R())
    # ex3) bc = [list(R()) for _ in range(m)]

# q = 2
# xdn = [[7, 2, 4], [12345, 67890, 2019]]

input2 = sys.stdin.readline
R = lambda: list(map(int, input2().split()))
q = int(input())
xdn = [R() for _ in range(q)]

# calculate
fac_list = modfactorial_list(MOD)
for x, d, n in xdn:
    if d == 0:
        print(modpow(x, n))
    else:
        x = moddiv(x, d)
        if x + n - 1 >= MOD:
            print(0)
        else:
            print(modmul(moddiv(fac_list[x+n-1], fac_list[x-1]),modpow(d, n)))
# output
# ex1) print(*l) => when l = [2, 5], printed 2 5 6

Buggy python submission (verdict: Runtime Error):
# for i, a in enumerate(iterable)
# q, mod = divmod(a, b)
# divmod(x, y) returns the tuple (x//y, x%y)
# manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
import sys
sys.setrecursionlimit(10**7)
from itertools import combinations, permutations # https://docs.python.org/ja/3/library/itertools.html
from math import factorial
def combinations_count(n, r):
    # faster than the following code
    # def combinations_count(n, r):
    #     return factorial(n) // (factorial(n - r) * factorial(r))
    if n - r < r: r = n - r
    if r == 0: return 1
    if r == 1: return n
    numerator = [n - r + k + 1 for k in range(r)]
    denominator = [k + 1 for k in range(r)]
    for p in range(2,r+1):
        pivot = denominator[p - 1]
        if pivot > 1:
            offset = (n - r) % p
            for k in range(p-1,r,p):
                numerator[k - offset] /= pivot
                denominator[k] /= pivot
    result = 1
    for k in range(r):
        if numerator[k] > 1:
            result *= int(numerator[k])
    return result
def combination_with_repetition_count(n, r):
    return combinations_count(n + r - 1, r)
from collections import deque, Counter # https://docs.python.org/ja/3/library/collections.html#collections.deque
from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
from functools import reduce
from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
def gcds(numbers):
    return reduce(gcd, numbers)
def lcm(x, y):
    return (x * y) // gcd(x, y)
def lcms(numbers):
    return reduce(lcm, numbers, 1)

# first create factorial_list!
# fac_list = mod_factorial_list(n)
INF = 10 ** 18
MOD = 10 ** 6 + 3
def modpow(a, n, p = MOD):
    if   n == 0     : return 1
    elif n == 1     : return a % p
    elif n % 2 == 0 : return modpow(a, n//2, p)**2 % p
    else            : return modpow(a, n//2, p)**2 * a % p
def modinv(a, p = MOD):
    # evaluate reciprocal using Fermat's little theorem:
    # a**(p-1) is identical to 1 (mod p) when a and p is coprime
    return modpow(a, p-2, p)
def modinv_list(n, p = MOD):
    if n <= 1:
        return [0,1][:n+1]
    else:
        inv_t = [0,1]
        for i in range(2, n+1):
            inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
        return inv_t
def modfactorial_list(n, p = MOD):
    if n == 0:
        return [1]
    else:
        l = [1]
        tmp = 1
        for i in range(1, n+1):
            tmp = tmp * i % p
            l.append(tmp)
        return l
def modcomb(n, k, fac_list = [], p = MOD):
    # fac_list = modfactorial_list(100)
    # print(modcomb(100, 5, modfactorial_list(100)))
    from math import factorial
    if n < 0 or k < 0 or n < k: return 0
    if n == 0 or k == 0: return 1
    if len(fac_list) <= n:
        a = factorial(n) % p
        b = factorial(k) % p
        c = factorial(n-k) % p
    else:
        a = fac_list[n]
        b = fac_list[k]
        c = fac_list[n-k]
    return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
def modadd(a, b, p = MOD):
    return (a + b) % MOD
def modsub(a, b, p = MOD):
    return (a - b) % p
def modmul(a, b, p = MOD):
    return ((a % p) * (b % p)) % p
def moddiv(a, b, p = MOD):
    return modmul(a, modpow(b, p-2, p))

# set the inputs
    # open(0).read() is a convenient method:
    # ex) n, m, *x = map(int, open(0).read().split())
    #     min(x[::2]) - max(x[1::2])
    # ex2) *x, = map(int, open(0).read().split())
    #     don't forget to add comma after *x if only one variable is used
    # R = lambda: list(map(int, input2().split()))
    # old_R = lambda: list(map(int, input().split()))
    # ex1) n, k = R()
    # ex2) v = list(R())
    # ex3) bc = [list(R()) for _ in range(m)]

# q = 2
# xdn = [[7, 2, 4], [12345, 67890, 2019]]

input2 = sys.stdin.readline
R = lambda: list(map(int, input2().split()))
q = R()
xdn = [R() for _ in range(q)]

# calculate
fac_list = modfactorial_list(MOD)
for (x, d, n) in xdn:
    x = moddiv(x, d)
    print(modmul(moddiv(fac_list[x+n-1], fac_list[x-1]),modpow(d, n)))
# output
# ex1) print(*l) => when l = [2, 5], printed 2 5 6


Fixed submission:
# for i, a in enumerate(iterable)
# q, mod = divmod(a, b)
# divmod(x, y) returns the tuple (x//y, x%y)
# manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
import sys
sys.setrecursionlimit(10**7)
from itertools import combinations, permutations # https://docs.python.org/ja/3/library/itertools.html
from math import factorial
def combinations_count(n, r):
    # faster than the following code
    # def combinations_count(n, r):
    #     return factorial(n) // (factorial(n - r) * factorial(r))
    if n - r < r: r = n - r
    if r == 0: return 1
    if r == 1: return n
    numerator = [n - r + k + 1 for k in range(r)]
    denominator = [k + 1 for k in range(r)]
    for p in range(2,r+1):
        pivot = denominator[p - 1]
        if pivot > 1:
            offset = (n - r) % p
            for k in range(p-1,r,p):
                numerator[k - offset] /= pivot
                denominator[k] /= pivot
    result = 1
    for k in range(r):
        if numerator[k] > 1:
            result *= int(numerator[k])
    return result
def combination_with_repetition_count(n, r):
    return combinations_count(n + r - 1, r)
from collections import deque, Counter # https://docs.python.org/ja/3/library/collections.html#collections.deque
from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
from functools import reduce
from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
def gcds(numbers):
    return reduce(gcd, numbers)
def lcm(x, y):
    return (x * y) // gcd(x, y)
def lcms(numbers):
    return reduce(lcm, numbers, 1)

# first create factorial_list!
# fac_list = mod_factorial_list(n)
INF = 10 ** 18
MOD = 10 ** 6 + 3
def modpow(a, n, p = MOD):
    if   n == 0     : return 1
    elif n == 1     : return a % p
    elif n % 2 == 0 : return modpow(a, n//2, p)**2 % p
    else            : return modpow(a, n//2, p)**2 * a % p
def modinv(a, p = MOD):
    # evaluate reciprocal using Fermat's little theorem:
    # a**(p-1) is identical to 1 (mod p) when a and p is coprime
    return modpow(a, p-2, p)
def modinv_list(n, p = MOD):
    if n <= 1:
        return [0,1][:n+1]
    else:
        inv_t = [0,1]
        for i in range(2, n+1):
            inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
        return inv_t
def modfactorial_list(n, p = MOD):
    if n == 0:
        return [1]
    else:
        l = [0] * (n+1)
        tmp = 1
        for i in range(1, n+1):
            tmp = tmp * i % p
            l[i] = tmp
        return l
def modcomb(n, k, fac_list = [], p = MOD):
    # fac_list = modfactorial_list(100)
    # print(modcomb(100, 5, modfactorial_list(100)))
    from math import factorial
    if n < 0 or k < 0 or n < k: return 0
    if n == 0 or k == 0: return 1
    if len(fac_list) <= n:
        a = factorial(n) % p
        b = factorial(k) % p
        c = factorial(n-k) % p
    else:
        a = fac_list[n]
        b = fac_list[k]
        c = fac_list[n-k]
    return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
def modadd(a, b, p = MOD):
    return (a + b) % MOD
def modsub(a, b, p = MOD):
    return (a - b) % p
def modmul(a, b, p = MOD):
    return ((a % p) * (b % p)) % p
def moddiv(a, b, p = MOD):
    return modmul(a, modpow(b, p-2, p))

# set the inputs
    # open(0).read() is a convenient method:
    # ex) n, m, *x = map(int, open(0).read().split())
    #     min(x[::2]) - max(x[1::2])
    # ex2) *x, = map(int, open(0).read().split())
    #     don't forget to add comma after *x if only one variable is used
    # R = lambda: list(map(int, input2().split()))
    # old_R = lambda: list(map(int, input().split()))
    # ex1) n, k = R()
    # ex2) v = list(R())
    # ex3) bc = [list(R()) for _ in range(m)]

# q = 2
# xdn = [[7, 2, 4], [12345, 67890, 2019]]

input2 = sys.stdin.readline
R = lambda: list(map(int, input2().split()))
q = int(input())
xdn = [R() for _ in range(q)]

# calculate
fac_list = modfactorial_list(MOD)
for x, d, n in xdn:
    if d == 0:
        print(modpow(x, n))
    else:
        x = moddiv(x, d)
        if x + n - 1 >= MOD:
            print(0)
        else:
            print(modmul(moddiv(fac_list[x+n-1], fac_list[x-1]),modpow(d, n)))
# output
# ex1) print(*l) => when l = [2, 5], printed 2 5 6