#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const int MAX = 1005;
const int MAXK = 47;
int dp[MAXK][MAX][MAX];
int c[MAX][MAX];
int fact[MAX];
int ans[MAX][MAX];
int main() {
  c[0][0] = 1;
  for (int i = 1; i < MAX; i++) {
    for (int j = 0; j < MAX; j++) {
      c[i][j] = c[i - 1][j];
      if (j != 0) {
        c[i][j] += c[i - 1][j - 1];
      }
      if (c[i][j] > MOD) {
        c[i][j] %= MOD;
      }
    }
  }
  fact[0] = 1;
  for (int i = 1; i < MAX; i++) {
    fact[i] = ((long long)fact[i - 1] * i) % MOD;
  }
  dp[1][0][0] = 1;
  for (int j = 0; j < MAX; j++) {
    dp[0][j][0] = 1;
  }
  for (int i = 1; i < MAXK; i++) {
    for (int j = 1; j < MAX; j++) {
      for (int k = 0; k < MAX; k++) {
        dp[i][j][k] = dp[i][j - 1][k];
        if (k - j >= 0) {
          dp[i][j][k] += dp[i - 1][j - 1][k - j];
        }
        dp[i][j][k] %= MOD;
      }
    }
  }
  for (int i = 1; i < MAX; i++) {
    for (int j = 1; j < MAXK; j++) {
      for (int k = 0; k < i - j + 1; k++) {
        long long add = (long long)fact[j] * c[i - k][j];
        add %= MOD;
        add = (add * (dp[j][MAX - 1][k])) % MOD;
        ans[i][j] += add;
        if (ans[i][j] > MOD) {
          ans[i][j] %= MOD;
        }
      }
    }
  }
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    printf("%d\n", ans[a][b]);
  }
}

### 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 int MOD = 1000000007;
const int MAX = 1005;
const int MAXK = 47;
int dp[MAXK][MAX][MAX];
int c[MAX][MAX];
int fact[MAX];
int ans[MAX][MAX];
int main() {
  c[0][0] = 1;
  for (int i = 1; i < MAX; i++) {
    for (int j = 0; j < MAX; j++) {
      c[i][j] = c[i - 1][j];
      if (j != 0) {
        c[i][j] += c[i - 1][j - 1];
      }
      if (c[i][j] > MOD) {
        c[i][j] %= MOD;
      }
    }
  }
  fact[0] = 1;
  for (int i = 1; i < MAX; i++) {
    fact[i] = ((long long)fact[i - 1] * i) % MOD;
  }
  dp[1][0][0] = 1;
  for (int j = 0; j < MAX; j++) {
    dp[0][j][0] = 1;
  }
  for (int i = 1; i < MAXK; i++) {
    for (int j = 1; j < MAX; j++) {
      for (int k = 0; k < MAX; k++) {
        dp[i][j][k] = dp[i][j - 1][k];
        if (k - j >= 0) {
          dp[i][j][k] += dp[i - 1][j - 1][k - j];
        }
        dp[i][j][k] %= MOD;
      }
    }
  }
  for (int i = 1; i < MAX; i++) {
    for (int j = 1; j < MAXK; j++) {
      for (int k = 0; k < i - j + 1; k++) {
        long long add = (long long)fact[j] * c[i - k][j];
        add %= MOD;
        add = (add * (dp[j][MAX - 1][k])) % MOD;
        ans[i][j] += add;
        if (ans[i][j] > MOD) {
          ans[i][j] %= MOD;
        }
      }
    }
  }
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    printf("%d\n", ans[a][b]);
  }
}

```