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: Wrong Answer)

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
typedef long long ll;
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#define ios ios_base::sync_with_stdio(0) ; cin.tie(0) ; cout.tie(0)
#define so sizeof
#define pb push_back
#define cl clear() ;
#define vl vector<ll>
#define sz size()
#define len length()
#define el endl;cout.flush()
#define be begin()
#define fi first
#define se second
#define br break
#define en end()
#define ro return 0
#define br break
#define eb emplace_back
#define con continue
#define ms(x) memset(x , 0, so x)
#define all(x) (x).be , (x).en
#define rall(x) (x).rbegin() , (x).rend()
#define acc(x) accumulate((x).be , (x).en , 0ll)
#define forn(i,a,b) for(ll i=a;i<=b;++i)
#define rng_58 mt19937 rng(chrono::steady_clock::now().time_since_epoch().count())
#define vll vector<pair<ll,ll> >
#define pll pair<ll,ll>
#define plll pair<ll,pair<ll,ll>>
#define plvl pair<pair<ll,ll> ,ll>
#define mp make_pair
#define trace2(a,b) cerr <<"a is " << a << " b is " << b << el;
#define trace3(a,b,c) cerr <<"a is " << a << " b is " << b << " c is " << c << el;
#define trace4(a,b,c,d) cerr <<"a is " << a << " b is " << b << " c is " << c <<" d is " << d << el ;
#define trace5(a,b,c,d,e) cerr <<"a is " << a << " b is " << b << " c is " << c <<" d is " << d << " e is " << e << el;
#define trace6(a,b,c,d,e,f) cerr <<"a is " << a << " b is " << b << " c is " << c <<" d is " << d << " e is " << e << " f is " << f << el ;
const ll k = 1000003;
#define mod %
ll f[2*k+10] , inv[2*k+10] , fin[2*k+10];
void factorial()
{
    f[0] = 1 ;
    f[1] = 1 ;
    forn(i , 2 , 2*k+5){
       f[i] = (i*f[i-1]) mod k ;
    }
}
void inverse()
{
    inv[0] = inv[1] = 1 ;
    forn(i , 2 , 2*k+5){
       inv[i] = (inv[k mod i]*(k-k/i)) mod k ;
    }
}
void factinverse()
{
    fin[0] = fin[1] = 1 ;
    forn(i , 2 , 2*k+5){
       fin[i] = (inv[i] * fin[i-1]) mod k ;
    }
}
ll exp(ll x , ll y)
{
    if(y == 0 ) return 1 ;
    if(y & 1){
        return (((x*exp(x,y/2)) mod k) * exp(x,y/2)) mod k;
    }
    else return (((exp(x,y/2)) mod k) * exp(x,y/2)) mod k;
}
int main()
{
    ios;
    factorial();
    inverse();
    factinverse();
    ll q;
    cin >> q ;
    while(q--){
        ll x , d , n ;
        cin >> x >> d >> n ;
        if(x == 0 ) {
            cout << 0 << el ;
            con ;
        }
        if(d==0){
            cout << exp(x , n ) ;
            con ;
        }
        else {
            ll dinv = exp(d , k-2) ;
            ll X = (x*dinv) mod k ;
            if(X + n -1 >= k ) {
                cout << 0 << el ;
                con ;
            }
            ll ans;
            if(X>=1){
                ans = (f[X+n-1]*fin[X-1]) mod k ;
            }
            else ans = (f[X+n-1]) mod k ;
            ll dn = exp(d , n ) ;
            ans = (ans * dn) mod k ;
            cout << ans << el ;
            con ;
        }
    }
}