#include <bits/stdc++.h>
using namespace std;
#define int long long
#define p (int)(1e6 + 3)

int MOD(int a){
	a %= p;
	return a >= 0 ? a : a + p;
}

int power(int a, int N){
	if(N == 0){
		return 1;
	}
	else if(N % 2 == 0){
		return power(MOD(a * a), N / 2);
	}
	else{
		return MOD(a * power(a, N - 1));
	}
}

int inverse(int a){
	return power(MOD(a), p - 2);
}

int *fact;

int facto(int n){
	if(n >= p){
		return 0;
	}
	else{
		return fact[n];
	}
}

signed main(){
	int Q, q, i;
	fact = (int *)malloc(sizeof(int) * (p + 1));
	fact[0] = 1;
	for(i = 1; i <= p; i++){
		fact[i] = MOD(i * fact[i - 1]);
	}
	scanf("%lld", &Q);
	for(q = 0; q < Q; q++){
		int x, d, n;
		scanf("%lld%lld%lld", &x, &d, &n);
		if(d == 0){
			printf("%lld\n", power(x, n));
			continue;
		}
		if(x == 0){
			printf("0\n");
			continue;
		}
		if((p - x) % d == 0){
			i = (p - x) / d;
			if(i < n){
				printf("0\n");
				continue;
			}
		}
		i = MOD(x * inverse(d));
		int ans = MOD(facto(i + n - 1) * power(d, i + n - 1));
		ans = MOD(ans * inverse(facto(i - 1) * power(d, i - 1)));
		printf("%lld\n", ans);
	}
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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.

Example

Input

2
7 2 4
12345 67890 2019


Output

9009
916936

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define p (int)(1e6 + 3)

int MOD(int a){
	a %= p;
	return a >= 0 ? a : a + p;
}

int power(int a, int N){
	if(N == 0){
		return 1;
	}
	else if(N % 2 == 0){
		return power(MOD(a * a), N / 2);
	}
	else{
		return MOD(a * power(a, N - 1));
	}
}

int inverse(int a){
	return power(MOD(a), p - 2);
}

int *fact;

int facto(int n){
	if(n >= p){
		return 0;
	}
	else{
		return fact[n];
	}
}

signed main(){
	int Q, q, i;
	fact = (int *)malloc(sizeof(int) * (p + 1));
	fact[0] = 1;
	for(i = 1; i <= p; i++){
		fact[i] = MOD(i * fact[i - 1]);
	}
	scanf("%lld", &Q);
	for(q = 0; q < Q; q++){
		int x, d, n;
		scanf("%lld%lld%lld", &x, &d, &n);
		if(d == 0){
			printf("%lld\n", power(x, n));
			continue;
		}
		if(x == 0){
			printf("0\n");
			continue;
		}
		if((p - x) % d == 0){
			i = (p - x) / d;
			if(i < n){
				printf("0\n");
				continue;
			}
		}
		i = MOD(x * inverse(d));
		int ans = MOD(facto(i + n - 1) * power(d, i + n - 1));
		ans = MOD(ans * inverse(facto(i - 1) * power(d, i - 1)));
		printf("%lld\n", ans);
	}
	return 0;
}
```