#include<bits/stdc++.h>
using namespace std;
#define int long long
const int M=1000003;
int fac[M],inv[M],x,d,n,T;
int ksm(int x,int y){
	int ans=1;
	for (;y;y>>=1,x=x*x%M)
		if (y&1)(ans*=x)%=M;
	return ans;	
}
signed main(){
	scanf("%lld",&T);
	fac[0]=inv[0]=1;
	for (int i=1;i<M;i++)fac[i]=i*fac[i-1]%M;
	for (int i=1;i<M;i++)inv[i]=ksm(fac[i],M-2);
	while (T--){
		scanf("%lld%lld%lld",&x,&d,&n);
		if (!d){
			printf("%lld\n",ksm(x,n));
			continue;
		}
		(x*=ksm(d,M-2))%=M;
		if (x+n-1>=M){
			puts("0");
			continue;
		}
		printf("%lld\n",ksm(d,n)*fac[x+n-1]%M*(x==0?1:inv[x-1])%M);
	}
	return 0;
}
### Prompt

Your challenge is to write 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 int long long
const int M=1000003;
int fac[M],inv[M],x,d,n,T;
int ksm(int x,int y){
	int ans=1;
	for (;y;y>>=1,x=x*x%M)
		if (y&1)(ans*=x)%=M;
	return ans;	
}
signed main(){
	scanf("%lld",&T);
	fac[0]=inv[0]=1;
	for (int i=1;i<M;i++)fac[i]=i*fac[i-1]%M;
	for (int i=1;i<M;i++)inv[i]=ksm(fac[i],M-2);
	while (T--){
		scanf("%lld%lld%lld",&x,&d,&n);
		if (!d){
			printf("%lld\n",ksm(x,n));
			continue;
		}
		(x*=ksm(d,M-2))%=M;
		if (x+n-1>=M){
			puts("0");
			continue;
		}
		printf("%lld\n",ksm(d,n)*fac[x+n-1]%M*(x==0?1:inv[x-1])%M);
	}
	return 0;
}
```