#include <bits/stdc++.h>
using namespace std;
int dp[51][1001][2];
int fact[51];
int ncr[1001][1001];
const int MOD = 1000000007;
void add(int &a, int b) {
  a += b;
  if (a < 0) a += MOD;
  if (a >= MOD) a -= MOD;
}
void mul(int &a, int b) { a = ((long long)a * b) % MOD; }
int ans[55][1005];
int main() {
  ncr[0][0] = 1;
  for (int i = 1; i <= 1000; i++) {
    for (int j = 0; j <= 1000; j++) {
      if (j > i)
        ncr[i][j] = 0;
      else {
        ncr[i][j] = ncr[i - 1][j];
        if (j) add(ncr[i][j], ncr[i - 1][j - 1]);
      }
    }
  }
  fact[0] = 1;
  for (int i = 1; i <= 50; i++) {
    fact[i] = fact[i - 1];
    mul(fact[i], i);
  }
  dp[0][0][0] = 1;
  dp[0][0][1] = 1;
  for (int k = 1; k <= 1000; k++) {
    for (int i = 1; i <= 50; i++) {
      for (int j = 1; j <= 1000; j++) {
        int b = (k % 2);
        dp[i][j][b] = dp[i][j][!b];
        if (j >= k) {
          add(dp[i][j][b], dp[i - 1][j - k][!b]);
        }
      }
    }
  }
  for (int i = 1; i <= 50; i++) {
    for (int j = 1; j <= 1000; j++) {
      mul(dp[i][j][0], fact[i]);
    }
  }
  for (int i = 1; i <= 50; i++) {
    for (int j = 1; j <= 1000; j++) {
      ans[i][j] = 0;
      for (int k = 1; k <= j; k++) {
        if (j - k + i >= 0) {
          add(ans[i][j], (long long)dp[i][k][0] * ncr[j - k + i][i] % MOD);
        }
      }
    }
  }
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    scanf("%d %d", &n, &k);
    if (k > 50)
      printf("0\n");
    else {
      printf("%d\n", ans[k][n]);
    }
  }
}

### Prompt

Your challenge is to write a cpp solution to 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;
int dp[51][1001][2];
int fact[51];
int ncr[1001][1001];
const int MOD = 1000000007;
void add(int &a, int b) {
  a += b;
  if (a < 0) a += MOD;
  if (a >= MOD) a -= MOD;
}
void mul(int &a, int b) { a = ((long long)a * b) % MOD; }
int ans[55][1005];
int main() {
  ncr[0][0] = 1;
  for (int i = 1; i <= 1000; i++) {
    for (int j = 0; j <= 1000; j++) {
      if (j > i)
        ncr[i][j] = 0;
      else {
        ncr[i][j] = ncr[i - 1][j];
        if (j) add(ncr[i][j], ncr[i - 1][j - 1]);
      }
    }
  }
  fact[0] = 1;
  for (int i = 1; i <= 50; i++) {
    fact[i] = fact[i - 1];
    mul(fact[i], i);
  }
  dp[0][0][0] = 1;
  dp[0][0][1] = 1;
  for (int k = 1; k <= 1000; k++) {
    for (int i = 1; i <= 50; i++) {
      for (int j = 1; j <= 1000; j++) {
        int b = (k % 2);
        dp[i][j][b] = dp[i][j][!b];
        if (j >= k) {
          add(dp[i][j][b], dp[i - 1][j - k][!b]);
        }
      }
    }
  }
  for (int i = 1; i <= 50; i++) {
    for (int j = 1; j <= 1000; j++) {
      mul(dp[i][j][0], fact[i]);
    }
  }
  for (int i = 1; i <= 50; i++) {
    for (int j = 1; j <= 1000; j++) {
      ans[i][j] = 0;
      for (int k = 1; k <= j; k++) {
        if (j - k + i >= 0) {
          add(ans[i][j], (long long)dp[i][k][0] * ncr[j - k + i][i] % MOD);
        }
      }
    }
  }
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    scanf("%d %d", &n, &k);
    if (k > 50)
      printf("0\n");
    else {
      printf("%d\n", ans[k][n]);
    }
  }
}

```