#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
int g[1001][51][1001];
long long ans[1001][51], prod[51], C[1001][1001], h[1001][51];
int main() {
  memset(g, 0, sizeof(g));
  memset(ans, 0, sizeof(ans));
  memset(C, 0, sizeof(C));
  prod[1] = 1;
  for (long long i = 2; i <= 50; i++) {
    prod[i] = prod[i - 1] * (long long)i % mod;
  }
  C[0][0] = 1;
  for (int i = 1; i <= 1000; i++)
    for (int j = 0; j <= i; j++) {
      C[i][j] = 0;
      if (j > 0) {
        C[i][j] += C[i - 1][j - 1];
      }
      if (j < i) {
        C[i][j] += C[i - 1][j];
      }
      if (C[i][j] >= mod) {
        C[i][j] -= mod;
      }
    }
  for (int i = 0; i <= 1000; i++) {
    g[0][0][i] = g[0][1][i] = 1;
  }
  for (int i = 1; i <= 1000; i++)
    for (int j = 1; j <= 50; j++) {
      for (int k = 1; k <= i; k++) {
        g[i][j][k] = g[i][j][k - 1] + g[i - k][j - 1][k - 1];
        if (g[i][j][k] >= mod) {
          g[i][j][k] -= mod;
        }
      }
      for (int k = i + 1; k <= 1000; k++) {
        g[i][j][k] = g[i][j][k - 1];
      }
    }
  for (int i = 1; i <= 1000; i++)
    for (int j = 1; j <= 50; j++) {
      h[i][j] = 0;
      if (i >= j) {
        for (int x = 0; x <= i - j; x++) {
          h[i][j] +=
              (long long)g[x][j][x] * prod[j] % mod * C[i - x - 1][j - 1] % mod;
          if (h[i][j] >= mod) {
            h[i][j] -= mod;
          }
        }
      }
      ans[i][j] = (ans[i - 1][j] + h[i][j]) % mod;
    }
  int test;
  scanf("%d\n", &test);
  while (test--) {
    int n, k;
    scanf("%d%d", &n, &k);
    if (k > 50) {
      printf("0\n");
      continue;
    }
    printf("%d\n", ans[n][k]);
  }
  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 long long mod = 1e9 + 7;
int g[1001][51][1001];
long long ans[1001][51], prod[51], C[1001][1001], h[1001][51];
int main() {
  memset(g, 0, sizeof(g));
  memset(ans, 0, sizeof(ans));
  memset(C, 0, sizeof(C));
  prod[1] = 1;
  for (long long i = 2; i <= 50; i++) {
    prod[i] = prod[i - 1] * (long long)i % mod;
  }
  C[0][0] = 1;
  for (int i = 1; i <= 1000; i++)
    for (int j = 0; j <= i; j++) {
      C[i][j] = 0;
      if (j > 0) {
        C[i][j] += C[i - 1][j - 1];
      }
      if (j < i) {
        C[i][j] += C[i - 1][j];
      }
      if (C[i][j] >= mod) {
        C[i][j] -= mod;
      }
    }
  for (int i = 0; i <= 1000; i++) {
    g[0][0][i] = g[0][1][i] = 1;
  }
  for (int i = 1; i <= 1000; i++)
    for (int j = 1; j <= 50; j++) {
      for (int k = 1; k <= i; k++) {
        g[i][j][k] = g[i][j][k - 1] + g[i - k][j - 1][k - 1];
        if (g[i][j][k] >= mod) {
          g[i][j][k] -= mod;
        }
      }
      for (int k = i + 1; k <= 1000; k++) {
        g[i][j][k] = g[i][j][k - 1];
      }
    }
  for (int i = 1; i <= 1000; i++)
    for (int j = 1; j <= 50; j++) {
      h[i][j] = 0;
      if (i >= j) {
        for (int x = 0; x <= i - j; x++) {
          h[i][j] +=
              (long long)g[x][j][x] * prod[j] % mod * C[i - x - 1][j - 1] % mod;
          if (h[i][j] >= mod) {
            h[i][j] -= mod;
          }
        }
      }
      ans[i][j] = (ans[i - 1][j] + h[i][j]) % mod;
    }
  int test;
  scanf("%d\n", &test);
  while (test--) {
    int n, k;
    scanf("%d%d", &n, &k);
    if (k > 50) {
      printf("0\n");
      continue;
    }
    printf("%d\n", ans[n][k]);
  }
  return 0;
}

```