#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using PII = pair<int, int>;
using VI = vector<int>;

const int mod = 1e6 + 3;
inline int add(int u, int v) {
  return u + v >= mod ? u + v - mod : u + v;
}
inline int mul(int u, int v) {
  return u * 1ll * v % mod;
}
int pwr(int a, int n) {
  int r = 1;
  for (; n > 0; n >>= 1) {
    if (n & 1) r = mul(r, a);
    a = mul(a, a);
  }
  return r;
}
inline int inv(int a) {
  return pwr(a, mod - 2);
}

int fact[mod];

int main() {
#ifdef LOCAL_EXEC
//	freopen("sample.in", "r", stdin);
//	freopen("sample.out", "w", stdout);
#else
  ios_base::sync_with_stdio(false); cin.tie(NULL);
#endif

  fact[0] = 1;
  for (int i = 1; i < mod; i++) {
    fact[i] = mul(fact[i - 1], i);
  }

  int q; cin >> q;
  while (q--) {
    int x, d, n; cin >> x >> d >> n;
    int ans = 0;
    if (d == 0) {
      ans = pwr(x, n);
    } else {
      int y = mul(x, inv(d));
      if (y + n - 1 < mod) {
        ans = mul(pwr(d, n), mul(fact[y + n - 1], inv(fact[y - 1])));
      }
    }
    cout << ans << endl;
  }

  return 0;
}

### 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;
using ll = long long;
using PII = pair<int, int>;
using VI = vector<int>;

const int mod = 1e6 + 3;
inline int add(int u, int v) {
  return u + v >= mod ? u + v - mod : u + v;
}
inline int mul(int u, int v) {
  return u * 1ll * v % mod;
}
int pwr(int a, int n) {
  int r = 1;
  for (; n > 0; n >>= 1) {
    if (n & 1) r = mul(r, a);
    a = mul(a, a);
  }
  return r;
}
inline int inv(int a) {
  return pwr(a, mod - 2);
}

int fact[mod];

int main() {
#ifdef LOCAL_EXEC
//	freopen("sample.in", "r", stdin);
//	freopen("sample.out", "w", stdout);
#else
  ios_base::sync_with_stdio(false); cin.tie(NULL);
#endif

  fact[0] = 1;
  for (int i = 1; i < mod; i++) {
    fact[i] = mul(fact[i - 1], i);
  }

  int q; cin >> q;
  while (q--) {
    int x, d, n; cin >> x >> d >> n;
    int ans = 0;
    if (d == 0) {
      ans = pwr(x, n);
    } else {
      int y = mul(x, inv(d));
      if (y + n - 1 < mod) {
        ans = mul(pwr(d, n), mul(fact[y + n - 1], inv(fact[y - 1])));
      }
    }
    cout << ans << endl;
  }

  return 0;
}

```