#include <bits/stdc++.h>
using namespace std;
long long dp[1005][1005], sum[1005][1005], c[2 * 1005][2 * 1005];
long long res[55][1005], p[55];
int main() {
  dp[0][0] = 1;
  for (int i = 0; i < 1005; i++) sum[0][i] = 1;
  for (int i = 1; i <= 51; i++) {
    for (int j = 1005 - 1; j >= 1; j--) {
      for (int k = 1005 - 1; k >= 1; k--) {
        if ((j >= k && i == 1) || j > k)
          dp[j][k] = sum[j - k][k - 1];
        else
          dp[j][k] = 0;
        sum[j][k] = dp[j][k];
      }
      for (int k = 1; k < 1005; k++)
        sum[j][k] = (sum[j][k] + sum[j][k - 1]) % 1000000007;
      res[i][j] = sum[j][1005 - 1];
    }
    sum[0][0] = dp[0][0] = 0;
  }
  c[0][0] = c[1][1] = 1;
  for (int i = 2; i < 2 * 1005; i++)
    for (int j = 1; j <= i; j++)
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % 1000000007;
  p[0] = 1;
  for (int i = 1; i < 55; i++) p[i] = (p[i - 1] * i) % 1000000007;
  for (int i = 0; i <= 51; i++)
    for (int j = 1; j < 1005; j++)
      res[i][j] =
          (res[i][j - 1] + (res[i][j] * p[i]) % 1000000007) % 1000000007;
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, k;
    scanf("%d %d", &n, &k);
    if (k > 51)
      printf("0\n");
    else {
      long long ans = 0;
      for (int i = k; i <= n; i++)
        ans = (ans + (res[k][i] * c[n - i + k][k]) % 1000000007) % 1000000007;
      printf("%I64d\n", ans);
    }
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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;
long long dp[1005][1005], sum[1005][1005], c[2 * 1005][2 * 1005];
long long res[55][1005], p[55];
int main() {
  dp[0][0] = 1;
  for (int i = 0; i < 1005; i++) sum[0][i] = 1;
  for (int i = 1; i <= 51; i++) {
    for (int j = 1005 - 1; j >= 1; j--) {
      for (int k = 1005 - 1; k >= 1; k--) {
        if ((j >= k && i == 1) || j > k)
          dp[j][k] = sum[j - k][k - 1];
        else
          dp[j][k] = 0;
        sum[j][k] = dp[j][k];
      }
      for (int k = 1; k < 1005; k++)
        sum[j][k] = (sum[j][k] + sum[j][k - 1]) % 1000000007;
      res[i][j] = sum[j][1005 - 1];
    }
    sum[0][0] = dp[0][0] = 0;
  }
  c[0][0] = c[1][1] = 1;
  for (int i = 2; i < 2 * 1005; i++)
    for (int j = 1; j <= i; j++)
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % 1000000007;
  p[0] = 1;
  for (int i = 1; i < 55; i++) p[i] = (p[i - 1] * i) % 1000000007;
  for (int i = 0; i <= 51; i++)
    for (int j = 1; j < 1005; j++)
      res[i][j] =
          (res[i][j - 1] + (res[i][j] * p[i]) % 1000000007) % 1000000007;
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, k;
    scanf("%d %d", &n, &k);
    if (k > 51)
      printf("0\n");
    else {
      long long ans = 0;
      for (int i = k; i <= n; i++)
        ans = (ans + (res[k][i] * c[n - i + k][k]) % 1000000007) % 1000000007;
      printf("%I64d\n", ans);
    }
  }
  return 0;
}

```