#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007;
int C[2001][2001];
int F[48][1011][1011];
long long ways[1011][1011];
int MAIN() {
  memset(C, 0, sizeof(C));
  for (int i = 0; i <= 2000; i++) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; j++)
      C[i][j] = ((long long)C[i - 1][j - 1] + C[i - 1][j]) % mod;
  }
  memset(F, 0, sizeof(F));
  memset(ways, 0, sizeof(ways));
  F[0][0][0] = 1;
  for (int i = 0; i <= 47; i++)
    for (int j = 0; j <= 1000; j++)
      for (int k = 0; k <= 1000; k++)
        if (F[i][j][k] > 0) {
          if (j + 1 <= 1000)
            F[i][j + 1][k] = (F[i][j + 1][k] + F[i][j][k]) % mod;
          if (j + 1 <= 1001 && j + k <= 1000 && i + 1 <= 47)
            F[i + 1][j + 1][k + j] =
                (F[i + 1][j + 1][k + j] + F[i][j][k]) % mod;
          ways[i][k] = F[i][j][k];
        }
  int T;
  cin >> T;
  int N, K;
  while (T--) {
    cin >> N >> K;
    N -= K;
    if (K >= 47 || N < 0)
      cout << 0 << endl;
    else {
      long long ret = 0;
      for (int len = 0; len <= N; len++) {
        long long now = ways[K][len];
        now *= C[(N - len) + K][K];
        ret += now;
        ret %= mod;
      }
      for (int i = 1; i <= K; i++) ret = (ret * i) % mod;
      cout << ret << endl;
    }
  }
  return 0;
}
int main() {
  ios ::sync_with_stdio(false);
  cout << fixed << setprecision(16);
  return MAIN();
}

### Prompt

In cpp, your task is to solve 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 mod = 1000000007;
int C[2001][2001];
int F[48][1011][1011];
long long ways[1011][1011];
int MAIN() {
  memset(C, 0, sizeof(C));
  for (int i = 0; i <= 2000; i++) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; j++)
      C[i][j] = ((long long)C[i - 1][j - 1] + C[i - 1][j]) % mod;
  }
  memset(F, 0, sizeof(F));
  memset(ways, 0, sizeof(ways));
  F[0][0][0] = 1;
  for (int i = 0; i <= 47; i++)
    for (int j = 0; j <= 1000; j++)
      for (int k = 0; k <= 1000; k++)
        if (F[i][j][k] > 0) {
          if (j + 1 <= 1000)
            F[i][j + 1][k] = (F[i][j + 1][k] + F[i][j][k]) % mod;
          if (j + 1 <= 1001 && j + k <= 1000 && i + 1 <= 47)
            F[i + 1][j + 1][k + j] =
                (F[i + 1][j + 1][k + j] + F[i][j][k]) % mod;
          ways[i][k] = F[i][j][k];
        }
  int T;
  cin >> T;
  int N, K;
  while (T--) {
    cin >> N >> K;
    N -= K;
    if (K >= 47 || N < 0)
      cout << 0 << endl;
    else {
      long long ret = 0;
      for (int len = 0; len <= N; len++) {
        long long now = ways[K][len];
        now *= C[(N - len) + K][K];
        ret += now;
        ret %= mod;
      }
      for (int i = 1; i <= K; i++) ret = (ret * i) % mod;
      cout << ret << endl;
    }
  }
  return 0;
}
int main() {
  ios ::sync_with_stdio(false);
  cout << fixed << setprecision(16);
  return MAIN();
}

```