#include<bits/stdc++.h>
#define mod 1000003
#define maxn 1000010
#define ll long long 
using namespace std;
int pre[maxn];
int calc(int a, int k)
{
	int ans = 1;
	while (k)
	{
		if (k % 2) ans = (ll)ans * a % mod;
		a = (ll)a * a % mod;
		k /= 2;
	}
	return ans;
}
int main()
{
	pre[0] = 1;
	pre[1] = 1;
	for (int i = 2; i <= mod - 1; i++)
		pre[i] = (ll)pre[i - 1] * i % mod;
	int T;
	scanf("%d", &T);
	while (T--)
	{
		int x, d, n, res;
		scanf("%d%d%d", &x, &d, &n);
		if (d == 0) {printf("%d\n", calc(x, n)); continue;}
		res = calc(d, n);
		x = (ll)x * calc(d, mod - 2) % mod;
		int l = x, r = x + n - 1;
		if (r >= mod || x == 0) {puts("0"); continue;}
		res = (ll)res * pre[r] % mod * calc(pre[l - 1], mod - 2) % mod;
		printf("%d\n", res);
	}
	return 0;
}
### Prompt

Create a solution in CPP for the following problem:

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>
#define mod 1000003
#define maxn 1000010
#define ll long long 
using namespace std;
int pre[maxn];
int calc(int a, int k)
{
	int ans = 1;
	while (k)
	{
		if (k % 2) ans = (ll)ans * a % mod;
		a = (ll)a * a % mod;
		k /= 2;
	}
	return ans;
}
int main()
{
	pre[0] = 1;
	pre[1] = 1;
	for (int i = 2; i <= mod - 1; i++)
		pre[i] = (ll)pre[i - 1] * i % mod;
	int T;
	scanf("%d", &T);
	while (T--)
	{
		int x, d, n, res;
		scanf("%d%d%d", &x, &d, &n);
		if (d == 0) {printf("%d\n", calc(x, n)); continue;}
		res = calc(d, n);
		x = (ll)x * calc(d, mod - 2) % mod;
		int l = x, r = x + n - 1;
		if (r >= mod || x == 0) {puts("0"); continue;}
		res = (ll)res * pre[r] % mod * calc(pre[l - 1], mod - 2) % mod;
		printf("%d\n", res);
	}
	return 0;
}
```