#include <bits/stdc++.h>
using namespace std;
int n, K;
int tot[55][1009][1009], C[1009][1009];
long long F[55];
void init() {
  for (int i = 0; i <= 1000; ++i) tot[0][0][i] = 1;
  for (int i = 1; i < 50; ++i) {
    for (int j = (i + 1) * i / 2; j <= 1000; ++j) {
      for (int k = i; k <= 1000; ++k) {
        tot[i][j][k] =
            tot[i][j][k - 1] + (j - k >= 0 ? tot[i - 1][j - k][k - 1] : 0);
        tot[i][j][k] %= 1000000007;
      }
    }
  }
  C[0][0] = 1;
  for (int i = 1; i <= 1000; ++i) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j <= i; ++j) {
      C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
      C[i][j] %= 1000000007;
    }
  }
  F[1] = 1;
  for (int i = 2; i <= 50; ++i) F[i] = (F[i - 1] * (long long)i) % 1000000007;
}
int main() {
  init();
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, k;
    scanf("%d%d", &n, &k);
    long long ans = 0;
    for (int i = (k + 1) * k / 2; i <= n; ++i) {
      ans += ((long long)tot[k][i][1000] * C[n - i + k][k]) %
             (long long)1000000007;
      ans %= (long long)1000000007;
    }
    ans = (ans * F[k]) % (long long)1000000007;
    printf("%I64d\n", ans);
  }
}

### Prompt

Please create a solution in Cpp to the following problem:

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;
int n, K;
int tot[55][1009][1009], C[1009][1009];
long long F[55];
void init() {
  for (int i = 0; i <= 1000; ++i) tot[0][0][i] = 1;
  for (int i = 1; i < 50; ++i) {
    for (int j = (i + 1) * i / 2; j <= 1000; ++j) {
      for (int k = i; k <= 1000; ++k) {
        tot[i][j][k] =
            tot[i][j][k - 1] + (j - k >= 0 ? tot[i - 1][j - k][k - 1] : 0);
        tot[i][j][k] %= 1000000007;
      }
    }
  }
  C[0][0] = 1;
  for (int i = 1; i <= 1000; ++i) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j <= i; ++j) {
      C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
      C[i][j] %= 1000000007;
    }
  }
  F[1] = 1;
  for (int i = 2; i <= 50; ++i) F[i] = (F[i - 1] * (long long)i) % 1000000007;
}
int main() {
  init();
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, k;
    scanf("%d%d", &n, &k);
    long long ans = 0;
    for (int i = (k + 1) * k / 2; i <= n; ++i) {
      ans += ((long long)tot[k][i][1000] * C[n - i + k][k]) %
             (long long)1000000007;
      ans %= (long long)1000000007;
    }
    ans = (ans * F[k]) % (long long)1000000007;
    printf("%I64d\n", ans);
  }
}

```