Problem:
Score : 600 points

Problem Statement

Consider the following arithmetic progression with n terms:

x, x + d, x + 2d, \ldots, x + (n-1)d

What is the product of all terms in this sequence?
Compute the answer modulo 1\ 000\ 003.

You are given Q queries of this form.
In the i-th query, compute the answer in case x = x_i, d = d_i, n = n_i.

Constraints

1 \leq Q \leq 10^5

0 \leq x_i, d_i \leq 1\ 000\ 002

1 \leq n_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

Q
x_1 d_1 n_1
:
x_Q d_Q n_Q

Output

Print Q lines.

In the i-th line, print the answer for the i-th query.

Sample Input 1

2
7 2 4
12345 67890 2019

Sample Output 1

9009
916936

For the first query, the answer is 7 \times 9 \times 11 \times 13 = 9009.
Don't forget to compute the answer modulo 1\ 000\ 003.

Solution in PyPy3 (2.4.0) (judge verdict: Accepted)

# https://atcoder.jp/contests/m-solutions2019/submissions/5741430
from functools import reduce
from operator import mul
import sys

input = sys.stdin.readline

MOD = 10 ** 6 + 3


def inv(n):
    return pow(n, MOD - 2, MOD)


def mod_mul(x, y):
    return mul(x, y) % MOD


fact = [1]
for n in range(1, MOD):
    fact.append((fact[-1] * n) % MOD)
    # d>1 の数列を d=1 となる数列に変換する
    # MODの倍数を含む数列の積は0になるので,
    # MODの倍数を含まない区間の積が計算できればよい
    # そのような区間の右端はMOD-1であり,fact[MOD-1]が計算できればよい

inv_fact = [-1] * MOD
inv_fact[MOD - 1] = inv(fact[-1])
for n in range(MOD - 2, -1, -1):
    inv_fact[n] = ((inv_fact[n + 1] * (n + 1)) % MOD)
    # inv_factは,inv(n!)にn,n-1,...を掛けるイメージ
    
Q = int(input())
for _ in range(Q):
    x, d, n = map(int, input().split())

    if x == 0:
        print(0)
        continue

    if d == 0:
        print(pow(x, n, MOD))
        continue

    xd = (x * inv(d)) % MOD
    if MOD - xd < n:
        print(0)
        continue

    dn = pow(d, n, MOD)
    print(reduce(mod_mul, (fact[xd + n - 1], inv_fact[xd - 1], dn)))