#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define all(V) V.begin(),V.end()
#define MOD 1000003

ll modpow(ll x, ll k) {
	ll ans = 1;
	while (k) {
		if (k & 1)ans = ans * x % MOD;
		x = x * x % MOD;
		k /= 2;
	}
	return ans;
}

ll moddiv(ll x, ll y) {
	return x * modpow(y, MOD - 2) % MOD;
}

ll fact[1000005];

int main() {
	fact[0] = 1;
	for (ll i = 1;i < 1000005;i++) {
		fact[i] = fact[i - 1] * i % MOD;
	}
	int Q;
	cin >> Q;
	ll x, d, n, D, XD, d3, ans = 0;
	for (int i = 0;i < Q;i++) {
		cin >> x >> d >> n;
		if (d == 0) {
			cout << modpow(x, n) << endl;
			continue;
		}
		if (x == 0) {
			cout << 0 << endl;
			continue;
		}
		D = moddiv(x, d);
		if (D + n - 1 >= MOD)cout << 0 << endl;
		else 
			cout << moddiv(fact[D + n - 1], fact[D - 1]) * modpow(d, n) % MOD << endl;
	}
}
### Prompt

Your task is to create 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 ll long long
#define all(V) V.begin(),V.end()
#define MOD 1000003

ll modpow(ll x, ll k) {
	ll ans = 1;
	while (k) {
		if (k & 1)ans = ans * x % MOD;
		x = x * x % MOD;
		k /= 2;
	}
	return ans;
}

ll moddiv(ll x, ll y) {
	return x * modpow(y, MOD - 2) % MOD;
}

ll fact[1000005];

int main() {
	fact[0] = 1;
	for (ll i = 1;i < 1000005;i++) {
		fact[i] = fact[i - 1] * i % MOD;
	}
	int Q;
	cin >> Q;
	ll x, d, n, D, XD, d3, ans = 0;
	for (int i = 0;i < Q;i++) {
		cin >> x >> d >> n;
		if (d == 0) {
			cout << modpow(x, n) << endl;
			continue;
		}
		if (x == 0) {
			cout << 0 << endl;
			continue;
		}
		D = moddiv(x, d);
		if (D + n - 1 >= MOD)cout << 0 << endl;
		else 
			cout << moddiv(fact[D + n - 1], fact[D - 1]) * modpow(d, n) % MOD << endl;
	}
}
```