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: Accepted)

#include<bits/stdc++.h>
#include<algorithm>
#include<math.h>
#include<string>
#include<iostream>
#include<vector>
#include<cstdlib>
#include<stdio.h>
#include<cmath>
#include<limits>
#include<iomanip>

using namespace std;
 
template <class T> using V = vector<T>;
 
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
 
const long long INF = 1LL << 60;
 
using ll = long long;
using db = long double;
using st = string;
using ch = char;
using bl = bool;
using vll = V<ll>;
using vpll =V<pair<ll,ll>>;
using vst = V<st>;
using vdb = V<db>;
using vch = V<ch>;
using vbl = V<bl>;
using graph = V<V<int>>;
using pq = priority_queue<ll>;


 
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define rFOR(i,a,b) for(ll i=(a);i>=(b);i--)
#define oFOR(i,a,b) for(ll i=(a);i<(b);i+=2)
#define bgn begin()
#define en end()
#define SORT(a) sort((a).bgn,(a).en)
#define REV(a) reverse((a).bgn,(a).en)
#define M(a,b) max(a,b)
#define rM(a,b) min(a,b)
#define fi first
#define se second
#define sz size()
#define gcd(a,b) __gcd(a,b)
#define co(a) cout<<a<<endl;
#define pb(a) push_back(a);
  
 

ll Sum(ll n) {
ll m=0;
  while(n){
    m+=n%10;
    n/=10;
  }
  return m;
}
 
 
const int MAX = 510000;
const int MOD = 1000000007;
 
long long fac[MAX], finv[MAX], inv[MAX];
 
void Comuse() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
 
 
#define comuse Comuse()
 
ll combi(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

ll perm(int n,int k){
    if(n < k) return 0;
    if(n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] % MOD) % MOD;
}
 
 
ll lcm(ll a,ll b){
  ll n;
  n=a/gcd(a,b)*b;
  return n;
}


vll div(ll n){
  vll ret;
  for(ll i=1;i*i<=n;i++){
    if(n%i==0){
      ret.push_back(i);
      if(i*i!=n){
        ret.push_back(n/i);
      }
    }
  }
  SORT(ret);
  return (ret);
}

ll modpow(ll a,ll n,ll mod){
  ll ans=1;
  while(n>0){
    if(n&1){
      ans=ans*a%mod;
    }

    a=a*a%mod;
    
    n>>=1;
  }

  return ans;
}

void Solve();

signed main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout<<setprecision(20)<<fixed;
  Solve();
}


/****************************************\
| Thank you for viewing my code:)        |
| Written by RedSpica a.k.a. RanseMirage |
| Twitter:@asakaakasaka                  | 
\****************************************/
 

void Solve(){
  ll q;
    cin>>q;
  ll Mod=1000003;
  vll table(Mod);
  vll retable(Mod);
  
  table[0]=1;
  table[1]=1;
  FOR(i,2,Mod){
    table[i]=1;
    table[i]*=table[i-1]*i;
    table[i]%=Mod;
  }
  
  retable[0]=1;
  retable[1]=1;
  FOR(i,2,Mod){
    retable[i]=1;
    retable[i]*=modpow(i,Mod-2,Mod);
    retable[i]*=retable[i-1];
    retable[i]%=Mod;
  }
  
  FOR(i,0,q){
    ll ans=1;
    ll x,d,n;
      cin>>x>>d>>n;
    ll dm=modpow(d,Mod-2,Mod); 
    
    if(d==0){
      cout<<modpow(x,n,Mod)<<endl;
    }
    
    else if(n>=Mod){
      cout<<0<<endl;
    }  
    
    else if((x*dm+n-1)%Mod<=n-1){
      cout<<0<<endl;
    }
    else{
    ans*=table[(x*dm+n-1)%Mod];
    ans*=retable[(x*dm-1+Mod)%Mod];
    ans%=Mod;
    ans*=modpow(d,n,Mod);
    ans%=Mod;
    cout<<ans<<endl;
    }
  }
}