#include <bits/stdc++.h>
using namespace std;
using namespace std;
int T, k, n;
long long dp[2][1005][60];
long long C[1005][1005];
long long A[1005];
void ini() {
  A[1] = 1;
  for (int i = 2; i <= 1000; i++) A[i] = (A[i - 1] * i) % 1000000007;
  C[0][0] = 1;
  for (int i = 1; i <= 1000; i++) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; j++) {
      C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
      C[i][j] %= 1000000007;
    }
  }
}
int main() {
  ini();
  int cd = 1;
  int dd = 0;
  memset(dp, 0, sizeof(dp));
  dp[0][0][0] = 1;
  for (int i = 1; i <= 1000; i++) {
    for (int j = 0; j <= 1000; j++) {
      for (int u = 0; u <= 50; u++) {
        dp[cd][j][u] += dp[dd][j][u];
        dp[cd][j][u] %= 1000000007;
        if (j + i > 1000) continue;
        dp[cd][j + i][u + 1] += dp[dd][j][u];
        dp[cd][j + i][u + 1] %= 1000000007;
      }
    }
    cd = 1 - cd;
    dd = 1 - dd;
    for (int j = 0; j <= 1000; j++) {
      for (int u = 0; u <= 50; u++) {
        dp[cd][j][u] = 0;
      }
    }
  }
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d", &n, &k);
    if (k > 50) {
      printf("0\n");
      continue;
    }
    long long ans = 0;
    for (int j = 0; j <= n; j++) {
      if (dp[dd][j][k] != 0) {
        long long mx = dp[dd][j][k] * C[n - j + k][k];
        mx %= 1000000007;
        mx = (mx * A[k]) % 1000000007;
        ans = (ans + mx) % 1000000007;
      }
    }
    printf("%I64d\n", ans);
  }
  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;
using namespace std;
int T, k, n;
long long dp[2][1005][60];
long long C[1005][1005];
long long A[1005];
void ini() {
  A[1] = 1;
  for (int i = 2; i <= 1000; i++) A[i] = (A[i - 1] * i) % 1000000007;
  C[0][0] = 1;
  for (int i = 1; i <= 1000; i++) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; j++) {
      C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
      C[i][j] %= 1000000007;
    }
  }
}
int main() {
  ini();
  int cd = 1;
  int dd = 0;
  memset(dp, 0, sizeof(dp));
  dp[0][0][0] = 1;
  for (int i = 1; i <= 1000; i++) {
    for (int j = 0; j <= 1000; j++) {
      for (int u = 0; u <= 50; u++) {
        dp[cd][j][u] += dp[dd][j][u];
        dp[cd][j][u] %= 1000000007;
        if (j + i > 1000) continue;
        dp[cd][j + i][u + 1] += dp[dd][j][u];
        dp[cd][j + i][u + 1] %= 1000000007;
      }
    }
    cd = 1 - cd;
    dd = 1 - dd;
    for (int j = 0; j <= 1000; j++) {
      for (int u = 0; u <= 50; u++) {
        dp[cd][j][u] = 0;
      }
    }
  }
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d", &n, &k);
    if (k > 50) {
      printf("0\n");
      continue;
    }
    long long ans = 0;
    for (int j = 0; j <= n; j++) {
      if (dp[dd][j][k] != 0) {
        long long mx = dp[dd][j][k] * C[n - j + k][k];
        mx %= 1000000007;
        mx = (mx * A[k]) % 1000000007;
        ans = (ans + mx) % 1000000007;
      }
    }
    printf("%I64d\n", ans);
  }
  return 0;
};

```