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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include<bits/stdc++.h>

#define debug(x) cerr << #x << ": " << x << '\n'
#define debugArray(x,n) for(long long hoge = 0; (hoge) < (n); ++ (hoge)) cerr << #x << "[" << hoge << "]: " << x[hoge] << '\n'
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef tuple<ll,ll> Pll;
typedef vector<ll> vll;
const ll INF = LLONG_MAX/10;
const ll MOD = 1e9+7;


template <ll mod>
struct ModCombination{
    ll N;
    vector<ll> fact,inv,finv;

    ModCombination(ll n):N(n){
        fact.assign(n+1, 1);
        inv.assign(n+1, 1);
        finv.assign(n+1,1);
        for(int i=1;i<n;i++){
            fact[i+1] = fact[i]*(i+1)%mod;
            inv[i+1] = inv[mod%(i+1)]*(mod-mod/(i+1))%mod;
            finv[i+1] = finv[i]*inv[i+1]%mod;
        }
    }

    ll Comb(ll n,ll k){
        if(k<0||n<k)return 0;
        return (fact[n]*finv[k]%mod)*finv[n-k]%mod;
    }
};

ll mod_pow(ll x,ll e){
  return e? e&1? x*mod_pow(x,e-1)%MOD:mod_pow(x*x%MOD,e/2):1;
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  ModCombination<MOD> c(MOD);
  ll Q;cin>>Q;
  for(ll q=0;q<Q;q++){
    ll x,d,n;cin>>x>>d>>n;
    if(n>=MOD){
      cout<<0<<endl;
      continue;
    }
    ll e=x*mod_pow(d,MOD-2)%MOD;
    cout<<mod_pow(d,n)*c.fact[e+n-1]%MOD*c.finv[e-1]<<endl;
  }
  return 0;
}