#include <bits/stdc++.h>
using namespace std;
long long dp[2000][2000];
long long sm[2000][2000];
long long fact[2000];
long long dp0[2000][2000];
int main() {
  for (int m = 0; m < 2000; m++) dp[m][0] = 1, sm[m][0] = m + 1;
  for (int k = 1; k <= 50; k++) {
    dp[0][k] = sm[0][k] = 0;
    dp0[0][k] = (k == 1);
    for (int m = 1; m < 2000; m++) {
      dp[m][k] = 0;
      if (k == 1) {
        if (m >= k) dp[m][k] += sm[m - k][k - 1];
      } else {
        if (m > k) dp[m][k] += sm[m - k - 1][k - 1];
      }
      if (m >= k) dp[m][k] += dp[m - k][k];
      sm[m][k] = sm[m - 1][k] + dp[m][k];
      dp0[m][k] = dp[m][k] + sm[m - 1][k - 1];
      if (k == 1) dp0[m][k]++;
      sm[m][k] %= 1000000007;
      dp[m][k] %= 1000000007;
      dp0[m][k] %= 1000000007;
    }
  }
  fact[1] = 1;
  for (int i = 2; i < 2000; i++) fact[i] = (fact[i - 1] * i) % 1000000007;
  int t;
  scanf("%d", &t);
  int n, k;
  for (int i = 0; i < t; i++) {
    scanf("%d %d", &n, &k);
    printf("%d\n", (dp0[n - 1][k] * fact[k]) % 1000000007);
  }
}

### 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;
long long dp[2000][2000];
long long sm[2000][2000];
long long fact[2000];
long long dp0[2000][2000];
int main() {
  for (int m = 0; m < 2000; m++) dp[m][0] = 1, sm[m][0] = m + 1;
  for (int k = 1; k <= 50; k++) {
    dp[0][k] = sm[0][k] = 0;
    dp0[0][k] = (k == 1);
    for (int m = 1; m < 2000; m++) {
      dp[m][k] = 0;
      if (k == 1) {
        if (m >= k) dp[m][k] += sm[m - k][k - 1];
      } else {
        if (m > k) dp[m][k] += sm[m - k - 1][k - 1];
      }
      if (m >= k) dp[m][k] += dp[m - k][k];
      sm[m][k] = sm[m - 1][k] + dp[m][k];
      dp0[m][k] = dp[m][k] + sm[m - 1][k - 1];
      if (k == 1) dp0[m][k]++;
      sm[m][k] %= 1000000007;
      dp[m][k] %= 1000000007;
      dp0[m][k] %= 1000000007;
    }
  }
  fact[1] = 1;
  for (int i = 2; i < 2000; i++) fact[i] = (fact[i - 1] * i) % 1000000007;
  int t;
  scanf("%d", &t);
  int n, k;
  for (int i = 0; i < t; i++) {
    scanf("%d %d", &n, &k);
    printf("%d\n", (dp0[n - 1][k] * fact[k]) % 1000000007);
  }
}

```