#include <bits/stdc++.h>
using namespace std;
const int mod = 1000003;
vector<long long> fact;
long long modpow(long long n, int k) {
	long long r = 1;
	while (k > 0) {
		if (k&1) (r *= n) %= mod;
		(n *= n) %= mod;
		k >>= 1;
	}
	return r;
}
long long modinv(int n) {
	return modpow(n, mod-2);
}
void solve() {
	long long X, D, N;
	cin >> X >> D >> N;
	if (D == 0) {
		cout << modpow(X, N) << endl;
	} else if (N >= mod) {
		cout << 0 << endl;
	} else {
		long long s = X * modinv(D) % mod;
		long long t = (X+(N-1)*D) * modinv(D) % mod;
		if (0 < s && s <= t) {
			cout << fact[t] * modinv(fact[s-1]) % mod * modpow(D, N) % mod << endl;
		} else {
			cout << 0 << endl;
		}
	}
}
int main() {
	fact = vector<long long>(mod);
	fact[0] = 1;
	for (unsigned i = 1; i < fact.size(); ++ i) fact[i] = fact[i-1] * i % mod;

	int Q;
	cin >> Q;
	for (int i = 0; i < Q; ++ i) {
		solve();
	}
}

### Prompt

In Cpp, your task is to solve 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;
const int mod = 1000003;
vector<long long> fact;
long long modpow(long long n, int k) {
	long long r = 1;
	while (k > 0) {
		if (k&1) (r *= n) %= mod;
		(n *= n) %= mod;
		k >>= 1;
	}
	return r;
}
long long modinv(int n) {
	return modpow(n, mod-2);
}
void solve() {
	long long X, D, N;
	cin >> X >> D >> N;
	if (D == 0) {
		cout << modpow(X, N) << endl;
	} else if (N >= mod) {
		cout << 0 << endl;
	} else {
		long long s = X * modinv(D) % mod;
		long long t = (X+(N-1)*D) * modinv(D) % mod;
		if (0 < s && s <= t) {
			cout << fact[t] * modinv(fact[s-1]) % mod * modpow(D, N) % mod << endl;
		} else {
			cout << 0 << endl;
		}
	}
}
int main() {
	fact = vector<long long>(mod);
	fact[0] = 1;
	for (unsigned i = 1; i < fact.size(); ++ i) fact[i] = fact[i-1] * i % mod;

	int Q;
	cin >> Q;
	for (int i = 0; i < Q; ++ i) {
		solve();
	}
}

```