#include <bits/stdc++.h>
#define int long long
#define pb push_back
#define mp make_pair
#define INF 0x3f3f3f3f
#define Inf 1000000000000000000LL
#define MOD 1000003
#define F first
#define S second
using namespace std;
typedef pair<int,int>pii;
int f[2000010];
int quick_pow(int x,int y){
	int out=1;
	int p=x%MOD;
	while(y){
		if(y&1)out=out*p%MOD;
		p=p*p%MOD;
		y>>=1;
	}
	return out;
}
signed main(){
	int T;
	cin>>T;
	f[0]=1;
	for(int i=1;i<=2000000;i++)f[i]=f[i-1]*i%MOD;
	while(T--){
		int x,d,n,ans=1;
		cin>>x>>d>>n;
		if(!d)ans=quick_pow(x,n);
		else if(n>MOD)ans=0;
		else{
			x=x*quick_pow(d,MOD-2)%MOD;
			ans=f[x+n-1]*quick_pow(f[x-1],MOD-2)%MOD*quick_pow(d,n)%MOD;
		}
		cout<<ans<<endl;
	}
	return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 int long long
#define pb push_back
#define mp make_pair
#define INF 0x3f3f3f3f
#define Inf 1000000000000000000LL
#define MOD 1000003
#define F first
#define S second
using namespace std;
typedef pair<int,int>pii;
int f[2000010];
int quick_pow(int x,int y){
	int out=1;
	int p=x%MOD;
	while(y){
		if(y&1)out=out*p%MOD;
		p=p*p%MOD;
		y>>=1;
	}
	return out;
}
signed main(){
	int T;
	cin>>T;
	f[0]=1;
	for(int i=1;i<=2000000;i++)f[i]=f[i-1]*i%MOD;
	while(T--){
		int x,d,n,ans=1;
		cin>>x>>d>>n;
		if(!d)ans=quick_pow(x,n);
		else if(n>MOD)ans=0;
		else{
			x=x*quick_pow(d,MOD-2)%MOD;
			ans=f[x+n-1]*quick_pow(f[x-1],MOD-2)%MOD*quick_pow(d,n)%MOD;
		}
		cout<<ans<<endl;
	}
	return 0;
}

```