#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000 + 10;
const int mod = 1000000007;
int fact[MAXN];
int C[MAXN][MAXN];
int dp[50][MAXN][MAXN];
int main() {
  for (int i = 0; i < MAXN; i++) {
    C[i][0] = 1;
    for (int j = 1; j <= i; j++)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mod;
  }
  for (int c = 0; c < 50; c++)
    for (int sum = 0; sum < MAXN; sum++)
      for (int m = 0; m < MAXN; m++) {
        if (c == 0)
          dp[c][sum][m] = (sum == 0);
        else if (m == 0)
          dp[c][sum][m] = (sum == 0) && (c == 1);
        else
          dp[c][sum][m] = (dp[c][sum][m - 1] +
                           ((sum >= m) ? (dp[c - 1][sum - m][m - 1]) : (0))) %
                          mod;
      }
  fact[0] = 1;
  for (int i = 1; i < MAXN; i++) fact[i] = (long long)fact[i - 1] * i % mod;
  int t;
  cin >> t;
  for (int o = 0; o < t; o++) {
    int k, n;
    cin >> n >> k;
    if (k >= 50)
      cout << 0 << endl;
    else {
      int ret = 0;
      for (int sum = 0; sum <= n - k; sum++)
        ret =
            (ret + (long long)dp[k][sum][sum] * fact[k] % mod * C[n - sum][k]) %
            mod;
      cout << ret << endl;
    }
  }
  return 0;
}

### Prompt

Generate 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;
const int MAXN = 1000 + 10;
const int mod = 1000000007;
int fact[MAXN];
int C[MAXN][MAXN];
int dp[50][MAXN][MAXN];
int main() {
  for (int i = 0; i < MAXN; i++) {
    C[i][0] = 1;
    for (int j = 1; j <= i; j++)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mod;
  }
  for (int c = 0; c < 50; c++)
    for (int sum = 0; sum < MAXN; sum++)
      for (int m = 0; m < MAXN; m++) {
        if (c == 0)
          dp[c][sum][m] = (sum == 0);
        else if (m == 0)
          dp[c][sum][m] = (sum == 0) && (c == 1);
        else
          dp[c][sum][m] = (dp[c][sum][m - 1] +
                           ((sum >= m) ? (dp[c - 1][sum - m][m - 1]) : (0))) %
                          mod;
      }
  fact[0] = 1;
  for (int i = 1; i < MAXN; i++) fact[i] = (long long)fact[i - 1] * i % mod;
  int t;
  cin >> t;
  for (int o = 0; o < t; o++) {
    int k, n;
    cin >> n >> k;
    if (k >= 50)
      cout << 0 << endl;
    else {
      int ret = 0;
      for (int sum = 0; sum <= n - k; sum++)
        ret =
            (ret + (long long)dp[k][sum][sum] * fact[k] % mod * C[n - sum][k]) %
            mod;
      cout << ret << endl;
    }
  }
  return 0;
}

```