#include<bits/stdc++.h>
using namespace std;
#define p 1000003
#define Maxn 1000010
int Q;
long long fac[Maxn], inv[Maxn], Cap, x, d, n, Res, Nah, Remain;

long long Pow(long long y, long long x)
{
    long long tmp=1;
    for (; x; x>>=1)
    {
        if (x&1) tmp=(tmp*y)%p;
        y=(y*y)%p;
    }
    return tmp;
}

int main()
{
	fac[0]=1; for(int i=1; i<=p; i++) fac[i]=fac[i-1]*i%p;
    inv[1]=1; for(int i=2; i<p; i++) inv[i]=(p-p/i)*inv[p%i]%p;
    inv[0]=1; for(int i=1; i<p; i++) inv[i]=inv[i]*inv[i-1]%p;
	scanf("%d", &Q);
	while (Q--)
	{
		scanf("%lld%lld%lld", &x, &d, &n);
		if (d==0)
		{
			printf("%lld\n", Pow(x, n));
			continue;
		}
		Cap=x*Pow(d, p-2)%p; Remain = Cap + n - 1; Res = 1;
		if (Remain>=p) { puts("0"); continue; }
		else
		{
			Res = ((Res * (fac[Cap+n-1] * inv[Cap-1] % p) % p) * Pow(d, n)) % p;
		}
		if (Res<0) Res+=p; 
		printf("%lld\n", Res);
	}
}
### Prompt

Please formulate a Cpp solution to 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>
using namespace std;
#define p 1000003
#define Maxn 1000010
int Q;
long long fac[Maxn], inv[Maxn], Cap, x, d, n, Res, Nah, Remain;

long long Pow(long long y, long long x)
{
    long long tmp=1;
    for (; x; x>>=1)
    {
        if (x&1) tmp=(tmp*y)%p;
        y=(y*y)%p;
    }
    return tmp;
}

int main()
{
	fac[0]=1; for(int i=1; i<=p; i++) fac[i]=fac[i-1]*i%p;
    inv[1]=1; for(int i=2; i<p; i++) inv[i]=(p-p/i)*inv[p%i]%p;
    inv[0]=1; for(int i=1; i<p; i++) inv[i]=inv[i]*inv[i-1]%p;
	scanf("%d", &Q);
	while (Q--)
	{
		scanf("%lld%lld%lld", &x, &d, &n);
		if (d==0)
		{
			printf("%lld\n", Pow(x, n));
			continue;
		}
		Cap=x*Pow(d, p-2)%p; Remain = Cap + n - 1; Res = 1;
		if (Remain>=p) { puts("0"); continue; }
		else
		{
			Res = ((Res * (fac[Cap+n-1] * inv[Cap-1] % p) % p) * Pow(d, n)) % p;
		}
		if (Res<0) Res+=p; 
		printf("%lld\n", Res);
	}
}
```