#include <bits/stdc++.h>
using namespace std;
long long int mod = 1000 * 1000 * 1000 + 7;
long long int b[2005][2005];
long long int p[1005][1005];
long long int fact[1020];
int main() {
  for (int i = (0); i < (2005); i++) b[i][0] = b[i][i] = 1;
  for (int i = (2); i < (2005); i++) {
    for (int j = (1); j < (i); j++)
      b[i][j] = (b[i - 1][j - 1] + b[i - 1][j]) % mod;
  }
  fact[0] = 1;
  for (int i = (1); i < (1005); i++) fact[i] = (i * 1LL * fact[i - 1]) % mod;
  p[0][0] = 1;
  for (int i = (1); i < (1005); i++) p[i][0] = 0;
  p[0][1] = 0;
  for (int i = (1); i < (1005); i++) p[i][1] = 1;
  for (int i = (2); i < (1005); i++) {
    for (int j = (1); j < (1005); j++) {
      long long int u1 = (j - i >= 0) ? p[j - i][i - 1] : 0;
      long long int u2 = (j - i >= 0) ? p[j - i][i] : 0;
      p[j][i] = (u1 + u2) % mod;
    }
  }
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, k;
    scanf("%d%d", &n, &k);
    long long int res = 0;
    for (int l = (k * (k + 1) / 2); l < (n + 1); l++) {
      long long int u = (n - l + k >= 0 ? b[n - l + k][k] : 0);
      res += (((p[l][k] * b[n - l + k][k]) % mod) * fact[k]) % mod;
      res = res % mod;
    }
    cout << res << endl;
  }
  return 0;
}

### Prompt

Please formulate 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 int mod = 1000 * 1000 * 1000 + 7;
long long int b[2005][2005];
long long int p[1005][1005];
long long int fact[1020];
int main() {
  for (int i = (0); i < (2005); i++) b[i][0] = b[i][i] = 1;
  for (int i = (2); i < (2005); i++) {
    for (int j = (1); j < (i); j++)
      b[i][j] = (b[i - 1][j - 1] + b[i - 1][j]) % mod;
  }
  fact[0] = 1;
  for (int i = (1); i < (1005); i++) fact[i] = (i * 1LL * fact[i - 1]) % mod;
  p[0][0] = 1;
  for (int i = (1); i < (1005); i++) p[i][0] = 0;
  p[0][1] = 0;
  for (int i = (1); i < (1005); i++) p[i][1] = 1;
  for (int i = (2); i < (1005); i++) {
    for (int j = (1); j < (1005); j++) {
      long long int u1 = (j - i >= 0) ? p[j - i][i - 1] : 0;
      long long int u2 = (j - i >= 0) ? p[j - i][i] : 0;
      p[j][i] = (u1 + u2) % mod;
    }
  }
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, k;
    scanf("%d%d", &n, &k);
    long long int res = 0;
    for (int l = (k * (k + 1) / 2); l < (n + 1); l++) {
      long long int u = (n - l + k >= 0 ? b[n - l + k][k] : 0);
      res += (((p[l][k] * b[n - l + k][k]) % mod) * fact[k]) % mod;
      res = res % mod;
    }
    cout << res << endl;
  }
  return 0;
}

```