#include <bits/stdc++.h>
using namespace std;
long long c[1024][64], fac[64], dp[1024][64];
long long calc(int n, int k) {
  long long ret = 0;
  if (k >= 64) return 0;
  for (int i = k * (k - 1) / 2; i <= n - k; ++i) {
    ret = (ret +
           (dp[i + k][k] * c[n - i][k] % 1000000007) * fac[k] % 1000000007) %
          1000000007;
  }
  return ret;
}
int main() {
  int t, n, k, i, j;
  for (fac[0] = i = 1; i < 64; ++i) fac[i] = fac[i - 1] * i % 1000000007;
  memset(c, 0, sizeof(c));
  for (c[0][0] = i = 1; i < 1024; ++i) {
    for (c[i][0] = j = 1; j < 64 && j <= i; ++j) {
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % 1000000007;
    }
  }
  memset(dp, 0, sizeof(dp));
  for (dp[1][1] = 1, i = 2; i < 1024; ++i) {
    for (dp[i][1] = 1, j = 2; j < 64 && j < i; ++j) {
      dp[i][j] = (dp[i - j][j - 1] + dp[i - j][j]) % 1000000007;
    }
  }
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &n, &k);
    printf("%I64d\n", calc(n, k));
  }
  return 0;
}

### Prompt

Your task is to create 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;
long long c[1024][64], fac[64], dp[1024][64];
long long calc(int n, int k) {
  long long ret = 0;
  if (k >= 64) return 0;
  for (int i = k * (k - 1) / 2; i <= n - k; ++i) {
    ret = (ret +
           (dp[i + k][k] * c[n - i][k] % 1000000007) * fac[k] % 1000000007) %
          1000000007;
  }
  return ret;
}
int main() {
  int t, n, k, i, j;
  for (fac[0] = i = 1; i < 64; ++i) fac[i] = fac[i - 1] * i % 1000000007;
  memset(c, 0, sizeof(c));
  for (c[0][0] = i = 1; i < 1024; ++i) {
    for (c[i][0] = j = 1; j < 64 && j <= i; ++j) {
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % 1000000007;
    }
  }
  memset(dp, 0, sizeof(dp));
  for (dp[1][1] = 1, i = 2; i < 1024; ++i) {
    for (dp[i][1] = 1, j = 2; j < 64 && j < i; ++j) {
      dp[i][j] = (dp[i - j][j - 1] + dp[i - j][j]) % 1000000007;
    }
  }
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &n, &k);
    printf("%I64d\n", calc(n, k));
  }
  return 0;
}

```