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 <iostream>
#include<cstdio>
#include<vector>
#include<string>
#include<algorithm>
#include<sstream>
#include<queue>
#include<set>
#include<map>
#include<bitset>
#include<stack>
#define i64 long long
#define vi32 vector<int>
#define vi64 vector<long long>
#define  ii pair<int,int>
const long long mod = 1e6+3, inf = 1e9, linf = 1e18;
using namespace std;
#define pb push_back
#define mp make_pair
#define N 1000003
using namespace std;
i64 f[N+3];
//int pre[N];
i64 fast(i64 a, i64 b)
{
	if (b == 0)return 1;
	i64 d = fast(a, b / 2);
	d = (d % mod * d % mod) % mod;
	if (b % 2)return a * d% mod;
	return d % mod;
}
i64 rev(i64 x)
{
	return fast(x,mod-2);
}
void slv(i64 x, i64 d, i64 n)
{
	if (x == 0)
	{
		cout << "0\n";
		return ;
	}
	if (d == 0)
	{
		cout << fast(x, n) << endl;
		return;
	}
	i64 cur = x * rev(d)%mod;
	if (cur + n - 1 >= mod)
	{
		cout << "0\n";
		return;
	}
	i64 res = f[cur + n - 1] * rev(f[cur - 1]) % mod;
	//cout << f[cur + n - 1] << " " << rev(f[cur - 1]) << endl;
	res = (fast(d, n)%mod * res % mod)%mod;
	cout << res << endl;
}
int main()
{
	f[0] = f[1] = 1;

	f[0] = 1;
	for (int i = 1; i < mod; i++)
		f[i] = (f[i-1]*i)%mod;
	//cout << f[500008];
	int q;
	cin >> q;
	while (q--)
	{
		i64 x, d, n;
		cin >> x >> d >> n;
		slv(x,d,n);
	}
}