#include <bits/stdc++.h>
using namespace std;
const int N = 1000;
const int mod = 1000000007;
int n, m, T, c[N * 2 + 3][N * 2 + 3], d[53][N + 3][N + 3], e[53],
    f[N + 3][N + 3];
int main() {
  cin >> T;
  for (int i = 0; i <= N * 2; i++) c[i][0] = 1;
  for (int i = 1; i <= N * 2; i++)
    for (int j = 1; j <= N * 2; j++) {
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
    }
  d[0][0][1] = 1;
  e[1] = 1;
  for (int i = 2; i <= 50; i++) e[i] = (1ll * e[i - 1] * i) % mod;
  for (int i = 0; i < 50; i++) {
    for (int j = 0; j <= N; j++) {
      for (int k = 1; k <= N; k++) {
        if (j + k <= N) d[i + 1][j + k][k + 1] += d[i][j][k];
        d[i][j][k + 1] += d[i][j][k];
        if (j + k <= N && d[i + 1][j + k][k + 1] >= mod)
          d[i + 1][j + k][k + 1] -= mod;
        if (d[i][j][k + 1] >= mod) d[i][j][k + 1] -= mod;
      }
    }
  }
  for (int i = 1; i <= 50; i++)
    for (int j = 1; j <= N; j++) {
      (d[i][j][N + 1] = (1ll * d[i][j][N + 1] * e[i]) % mod) %= mod;
    }
  for (int i = 1; i <= 50; i++)
    for (int l = 1; l <= N; l++) {
      for (int j = l; j <= N; j++) {
        f[i][j] += (1ll * d[i][l][N + 1] * c[j - l + i][i]) % mod;
        f[i][j] %= mod;
      }
    }
  for (int tt = 1; tt <= T; tt++) {
    scanf("%d%d", &n, &m);
    printf("%d\n", f[m][n]);
  }
  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;
const int N = 1000;
const int mod = 1000000007;
int n, m, T, c[N * 2 + 3][N * 2 + 3], d[53][N + 3][N + 3], e[53],
    f[N + 3][N + 3];
int main() {
  cin >> T;
  for (int i = 0; i <= N * 2; i++) c[i][0] = 1;
  for (int i = 1; i <= N * 2; i++)
    for (int j = 1; j <= N * 2; j++) {
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
    }
  d[0][0][1] = 1;
  e[1] = 1;
  for (int i = 2; i <= 50; i++) e[i] = (1ll * e[i - 1] * i) % mod;
  for (int i = 0; i < 50; i++) {
    for (int j = 0; j <= N; j++) {
      for (int k = 1; k <= N; k++) {
        if (j + k <= N) d[i + 1][j + k][k + 1] += d[i][j][k];
        d[i][j][k + 1] += d[i][j][k];
        if (j + k <= N && d[i + 1][j + k][k + 1] >= mod)
          d[i + 1][j + k][k + 1] -= mod;
        if (d[i][j][k + 1] >= mod) d[i][j][k + 1] -= mod;
      }
    }
  }
  for (int i = 1; i <= 50; i++)
    for (int j = 1; j <= N; j++) {
      (d[i][j][N + 1] = (1ll * d[i][j][N + 1] * e[i]) % mod) %= mod;
    }
  for (int i = 1; i <= 50; i++)
    for (int l = 1; l <= N; l++) {
      for (int j = l; j <= N; j++) {
        f[i][j] += (1ll * d[i][l][N + 1] * c[j - l + i][i]) % mod;
        f[i][j] %= mod;
      }
    }
  for (int tt = 1; tt <= T; tt++) {
    scanf("%d%d", &n, &m);
    printf("%d\n", f[m][n]);
  }
  return 0;
}

```