#include <bits/stdc++.h>
using namespace std;
const int N = 1000 + 10, K = 50 + 5, mod = 1e9 + 7;
int n = 1000, m, k = 50, ans, dp[K][N][N], c[N + K][K];
inline void add(int &a, int b) { a = (a + b) % mod; }
int main() {
  ios_base::sync_with_stdio(false);
  for (int i = 0; i <= n + k; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= min(k, i); j++)
      add(c[i][j], (c[i - 1][j] + c[i - 1][j - 1]) % mod);
  }
  dp[0][0][0] = 1;
  for (int num = 0; num <= k; num++)
    for (int sum = 0; sum <= n; sum++)
      for (int mx = 1; mx <= n; mx++)
        add(dp[num][sum][mx],
            dp[num][sum][mx - 1] +
                (num && sum >= mx ? dp[num - 1][sum - mx][mx - 1] : 0));
  int test;
  cin >> test;
  while (test--) {
    cin >> n >> k;
    if (k > 50) {
      cout << 0 << '\n';
      continue;
    }
    int ans = 0;
    for (int sum = k; sum <= n; sum++)
      add(ans, 1ll * dp[k][sum][sum] * c[n - sum + k][k] % mod);
    for (int i = 1; i <= k; i++) ans = 1ll * ans * i % mod;
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

The sequence of integer pairs (a1, b1), (a2, b2), ..., (ak, bk) is beautiful, if the following statements are fulfilled: 

  * 1 ≤ a1 ≤ b1 < a2 ≤ b2 < ... < ak ≤ bk ≤ n, where n is a given positive integer; 
  * all numbers b1 - a1, b2 - a2, ..., bk - ak are distinct. 



For the given number n find the number of beautiful sequences of length k. As the answer can be rather large, print the remainder after dividing it by 1000000007 (109 + 7).

Input

The first line contains integer t (1 ≤ t ≤ 2·105) — the number of the test data.

Each of the next t lines contains two integers n and k (1 ≤ k ≤ n ≤ 1000).

Output

For each test from the input print the answer to the problem modulo 1000000007 (109 + 7). Print the answers to the tests in the order in which the tests are given in the input.

Examples

Input

6
1 1
2 1
2 2
3 1
3 2
3 3


Output

1
3
0
6
2
0

Note

In the first test sample there is exactly one beautiful sequence: (1, 1).

In the second test sample, the following sequences are beautiful: 

  * (1, 1); 
  * (1, 2); 
  * (2, 2). 



In the fourth test sample, the following sequences are beautiful: 

  * (1, 1); 
  * (1, 2); 
  * (1, 3); 
  * (2, 2); 
  * (2, 3); 
  * (3, 3). 



In the fifth test sample, the following sequences are beautiful: 

  * (1, 1), (2, 3); 
  * (1, 2), (3, 3). 



In the third and sixth samples, there are no beautiful sequences.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1000 + 10, K = 50 + 5, mod = 1e9 + 7;
int n = 1000, m, k = 50, ans, dp[K][N][N], c[N + K][K];
inline void add(int &a, int b) { a = (a + b) % mod; }
int main() {
  ios_base::sync_with_stdio(false);
  for (int i = 0; i <= n + k; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= min(k, i); j++)
      add(c[i][j], (c[i - 1][j] + c[i - 1][j - 1]) % mod);
  }
  dp[0][0][0] = 1;
  for (int num = 0; num <= k; num++)
    for (int sum = 0; sum <= n; sum++)
      for (int mx = 1; mx <= n; mx++)
        add(dp[num][sum][mx],
            dp[num][sum][mx - 1] +
                (num && sum >= mx ? dp[num - 1][sum - mx][mx - 1] : 0));
  int test;
  cin >> test;
  while (test--) {
    cin >> n >> k;
    if (k > 50) {
      cout << 0 << '\n';
      continue;
    }
    int ans = 0;
    for (int sum = k; sum <= n; sum++)
      add(ans, 1ll * dp[k][sum][sum] * c[n - sum + k][k] % mod);
    for (int i = 1; i <= k; i++) ans = 1ll * ans * i % mod;
    cout << ans << '\n';
  }
  return 0;
}

```