#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 7;
long long M = 1e9 + 7;
long long f[N], dp[2][N][N], ch[N][N];
void solve() {
  long long n, k;
  cin >> n >> k;
  long long res = 0;
  for (long long m = 0; m <= n - k; m++) {
    long long tmp;
    tmp = ch[m + k - 1][k - 1];
    tmp = (tmp * dp[0][n - k - m][k]) % M;
    res = (res + tmp) % M;
  }
  cout << res << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ch[0][0] = 1;
  for (int i = 1; i < N; i++)
    for (int j = 0; j <= i; j++) {
      ch[i][j] = ch[i - 1][j];
      if (j > 0) ch[i][j] = (ch[i - 1][j] + ch[i - 1][j - 1]) % M;
    }
  f[0] = 1;
  for (long long i = 1; i < N; i++) f[i] = (i * f[i - 1]) % M;
  dp[0][0][0] = dp[0][0][1] = 1;
  for (int i = 1; i < N; i++)
    for (int j = 0; j < N; j++)
      for (int k = 0; k <= 50; k++) {
        dp[i % 2][j][k] = dp[(i - 1) % 2][j][k];
        if (j >= i && k > 0)
          dp[i % 2][j][k] =
              (dp[i % 2][j][k] + dp[(i - 1) % 2][j - i][k - 1] * k) % M;
      }
  for (int k = 0; k < N; k++)
    for (int j = 1; j < N; j++)
      dp[0][j][k] = (dp[0][j][k] + dp[0][j - 1][k]) % M;
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
const int N = 1e3 + 7;
long long M = 1e9 + 7;
long long f[N], dp[2][N][N], ch[N][N];
void solve() {
  long long n, k;
  cin >> n >> k;
  long long res = 0;
  for (long long m = 0; m <= n - k; m++) {
    long long tmp;
    tmp = ch[m + k - 1][k - 1];
    tmp = (tmp * dp[0][n - k - m][k]) % M;
    res = (res + tmp) % M;
  }
  cout << res << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ch[0][0] = 1;
  for (int i = 1; i < N; i++)
    for (int j = 0; j <= i; j++) {
      ch[i][j] = ch[i - 1][j];
      if (j > 0) ch[i][j] = (ch[i - 1][j] + ch[i - 1][j - 1]) % M;
    }
  f[0] = 1;
  for (long long i = 1; i < N; i++) f[i] = (i * f[i - 1]) % M;
  dp[0][0][0] = dp[0][0][1] = 1;
  for (int i = 1; i < N; i++)
    for (int j = 0; j < N; j++)
      for (int k = 0; k <= 50; k++) {
        dp[i % 2][j][k] = dp[(i - 1) % 2][j][k];
        if (j >= i && k > 0)
          dp[i % 2][j][k] =
              (dp[i % 2][j][k] + dp[(i - 1) % 2][j - i][k - 1] * k) % M;
      }
  for (int k = 0; k < N; k++)
    for (int j = 1; j < N; j++)
      dp[0][j][k] = (dp[0][j][k] + dp[0][j - 1][k]) % M;
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```