#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
typedef vector<vector<ll>> Mat;
#define ALL(v) (v).begin(),(v).end()
const ll INF = 1LL << 60;
const ll MOD = 1000003;

ll mod_pow(ll a, ll n) {
	if (n == 0) return 1;
	if (n % 2 == 0) {
		ll tmp = mod_pow(a, n / 2);
		return (tmp * tmp) % MOD;
	}
	return (a * mod_pow(a, n - 1)) % MOD;
}

vector<ll> fact;
vector<ll> inv_fact;

void comb_init(int n) {
	fact.resize(n + 1);
	inv_fact.resize(n + 1);
	fact[0] = inv_fact[0] = 1;
	for (int i = 1; i <= n; i++) {
		fact[i] = (fact[i - 1] * i) % MOD;
		inv_fact[i] = (inv_fact[i - 1] * mod_pow(i, MOD - 2)) % MOD;
	}
}

int main() {
	int Q;
	cin >> Q;
	comb_init(2 * MOD);
	for (int i = 0; i < Q; i++) {
		ll x, d, n;
		cin >> x >> d >> n;
		if (d == 0) {
			cout << mod_pow(x, n) << endl;
			continue;
		}
		if (n >= MOD) {
			cout << 0 << endl;
			continue;
		}
		(x *= mod_pow(d, MOD - 2)) %= MOD;
		ll ans = fact[x + (n - 1)];
		if (x > 0) (ans *= inv_fact[x - 1]) %= MOD;
		(ans *= mod_pow(d, n)) %= MOD;
		cout << ans << endl;
	}
}
### 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;
typedef long long ll;
typedef vector<vector<ll>> Mat;
#define ALL(v) (v).begin(),(v).end()
const ll INF = 1LL << 60;
const ll MOD = 1000003;

ll mod_pow(ll a, ll n) {
	if (n == 0) return 1;
	if (n % 2 == 0) {
		ll tmp = mod_pow(a, n / 2);
		return (tmp * tmp) % MOD;
	}
	return (a * mod_pow(a, n - 1)) % MOD;
}

vector<ll> fact;
vector<ll> inv_fact;

void comb_init(int n) {
	fact.resize(n + 1);
	inv_fact.resize(n + 1);
	fact[0] = inv_fact[0] = 1;
	for (int i = 1; i <= n; i++) {
		fact[i] = (fact[i - 1] * i) % MOD;
		inv_fact[i] = (inv_fact[i - 1] * mod_pow(i, MOD - 2)) % MOD;
	}
}

int main() {
	int Q;
	cin >> Q;
	comb_init(2 * MOD);
	for (int i = 0; i < Q; i++) {
		ll x, d, n;
		cin >> x >> d >> n;
		if (d == 0) {
			cout << mod_pow(x, n) << endl;
			continue;
		}
		if (n >= MOD) {
			cout << 0 << endl;
			continue;
		}
		(x *= mod_pow(d, MOD - 2)) %= MOD;
		ll ans = fact[x + (n - 1)];
		if (x > 0) (ans *= inv_fact[x - 1]) %= MOD;
		(ans *= mod_pow(d, n)) %= MOD;
		cout << ans << endl;
	}
}
```