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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <limits.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#include <algorithm>
#include <cassert>
#include <cfloat>
#include <complex>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <regex>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <vector>

using namespace std;

#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
#define rep(i, n) for (ll i = 0; i < (n); ++i)
#define repLRE(i, l, r) for (ll i = (l); i <= (r); ++i)
#define rrepLRE(i, l, r) for (ll i = (l); i >= (r); --i)
#define Sort(v) sort(v.begin(), v.end())
#define rSort(v) sort(v.rbegin(), v.rend())
#define Reverse(v) reverse(v.begin(), v.end())
#define Lower_bound(v, x) \
  distance(v.begin(), lower_bound(v.begin(), v.end(), x))
#define Upper_bound(v, x) \
  distance(v.begin(), upper_bound(v.begin(), v.end(), x))

using ll = long long;
using ull = unsigned long long;
using P = pair<ll, ll>;
using T = tuple<ll, ll, ll>;
using vll = vector<ll>;
using vP = vector<P>;
using vT = vector<T>;
using vvll = vector<vector<ll>>;
using vvP = vector<vector<P>>;
using dqll = deque<ll>;

ll dx[9] = {-1, 1, 0, 0, -1, -1, 1, 1, 0};
ll dy[9] = {0, 0, -1, 1, -1, 1, -1, 1, 0};

/* Macros reg. ends here */

const ll INF = 1LL << 50;

static const long long mod = 1000003;
struct mint {
  ll x;
  mint(ll x = 0) : x(x % mod) {}
  mint& operator+=(const mint a) {
    (x += a.x) %= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    (x += mod - a.x) %= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint ret(*this);
    return ret += a;
  }
  mint operator-(const mint a) const {
    mint ret(*this);
    return ret -= a;
  }
  mint operator*(const mint a) const {
    mint ret(*this);
    return ret *= a;
  }
  mint pow(ll t) const {
    if (t == 0) return mint(1);
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod - 2); }
  mint& operator/=(const mint a) { return (*this) *= a.inv(); }
  mint operator/(const mint a) const {
    mint ret(*this);
    return ret /= a;
  }
};
class modutils {
  vector<mint> fact, invfact;

 public:
  modutils(int n = 200005) : fact(n + 1), invfact(n + 1) {
    fact[0] = 1;
    for (int i = 1; i <= n; i++) fact[i] = fact[i - 1] * i;
    invfact[n] = fact[n].inv();
    for (int i = n; i >= 1; i--) invfact[i - 1] = invfact[i] * i;
  }
  mint pow(mint x, ll n) { return x.pow(n); }
  mint comb(ll n, ll k) {
    if (n < 0 || k < 0 || n < k) return 0;
    return fact[n] * invfact[k] * invfact[n - k];
  }
  mint perm(ll n, ll k) {
    if (n < 0 || k < 0 || n < k) return 0;
    return fact[n] * invfact[n - k];
  }
  mint fac(ll n) { return fact[n]; }
};
int main() {
  // ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(15);

  ll q;
  cin >> q;

  modutils mu(10001000);

  rep(i, q) {
    ll x, d, n;
    cin >> x >> d >> n;
    mint ans;
    if (n > mod || x == 0) {
      ans = 0;
    } else {
      if (d == 0) {
        ans = mint(x).pow(n);
      } else {
        mint k = mint(x) / mint(d);
        ll ilt = k.x + (n - 1);
        ans = mu.fac(ilt) / mu.fac(k.x - 1);
        ans *= mint(d).pow(n);
      }
    }
    cout << ans.x << endl;
  }

  return 0;
}