#include <bits/stdc++.h>
using namespace std;
int f, dp[2][1010][45], Fac[45], Ans[1010][1010];
void init() {
  int i, j, k;
  f = 1;
  for (i = 0; i < 1010; i++) {
    dp[f][i][0] = 1;
  }
  for (i = 0; i < 1010; i++) {
    f ^= 1;
    memset(dp[f], 0, sizeof(dp[f]));
    for (j = i + 1; j < 1010; j++) {
      for (k = 1; k < 45; k++) {
        if (dp[!f][j - i - 1][k - 1] == 0) {
          continue;
        }
        dp[f][j][k] = (dp[f][j][k] + dp[!f][j - i - 1][k - 1]) % 1000000007;
      }
    }
    for (j = 1; j < 1010; j++) {
      for (k = 0; k < 45; k++) {
        dp[f][j][k] = (dp[f][j][k] + dp[f][j - 1][k]) % 1000000007;
      }
    }
    for (j = 0; j < 1010; j++) {
      for (k = 0; k < 45; k++) {
        dp[f][j][k] = (dp[f][j][k] + dp[!f][j][k]) % 1000000007;
      }
    }
  }
  int Fac = 1;
  for (j = 1; j < 45; j++) {
    Fac = 1LL * Fac * j % 1000000007;
    for (i = 0; i < 1010; i++) {
      Ans[i][j] = 1LL * dp[f][i][j] * Fac % 1000000007;
    }
  }
}
int main() {
  init();
  int Case;
  scanf("%d", &Case);
  while (Case--) {
    int n, K;
    scanf("%d%d", &n, &K);
    printf("%d\n", Ans[n][K]);
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 f, dp[2][1010][45], Fac[45], Ans[1010][1010];
void init() {
  int i, j, k;
  f = 1;
  for (i = 0; i < 1010; i++) {
    dp[f][i][0] = 1;
  }
  for (i = 0; i < 1010; i++) {
    f ^= 1;
    memset(dp[f], 0, sizeof(dp[f]));
    for (j = i + 1; j < 1010; j++) {
      for (k = 1; k < 45; k++) {
        if (dp[!f][j - i - 1][k - 1] == 0) {
          continue;
        }
        dp[f][j][k] = (dp[f][j][k] + dp[!f][j - i - 1][k - 1]) % 1000000007;
      }
    }
    for (j = 1; j < 1010; j++) {
      for (k = 0; k < 45; k++) {
        dp[f][j][k] = (dp[f][j][k] + dp[f][j - 1][k]) % 1000000007;
      }
    }
    for (j = 0; j < 1010; j++) {
      for (k = 0; k < 45; k++) {
        dp[f][j][k] = (dp[f][j][k] + dp[!f][j][k]) % 1000000007;
      }
    }
  }
  int Fac = 1;
  for (j = 1; j < 45; j++) {
    Fac = 1LL * Fac * j % 1000000007;
    for (i = 0; i < 1010; i++) {
      Ans[i][j] = 1LL * dp[f][i][j] * Fac % 1000000007;
    }
  }
}
int main() {
  init();
  int Case;
  scanf("%d", &Case);
  while (Case--) {
    int n, K;
    scanf("%d%d", &n, &K);
    printf("%d\n", Ans[n][K]);
  }
  return 0;
}

```