#include <bits/stdc++.h>
using namespace std;
int dp[1005][1005][51];
int fact[1005];
const int mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(0);
  fact[0] = 1;
  for (int g = 1; g <= 1000; g++) fact[g] = (1LL * g * fact[g - 1]) % mod;
  for (int g = 0; g <= 1000; g++)
    for (int y = 0; y <= 1000; y++) dp[g][y][0] = 1;
  for (int g = 1; g <= 1000; g++) {
    for (int y = 0; y <= 1000; y++) {
      for (int t = 1; t <= 50; t++) {
        if (y == 0) {
          if (t == 1) {
            dp[g][y][t] = (1 + dp[g - 1][y][t]) % mod;
          }
          continue;
        }
        dp[g][y][t] = (dp[g - 1][y][t] + dp[g][y - 1][t] - dp[g - 1][y - 1][t]);
        dp[g][y][t] %= mod;
        dp[g][y][t] += mod;
        dp[g][y][t] %= mod;
        if (g - y > 0)
          dp[g][y][t] = (dp[g][y][t] + dp[g - y - 1][y - 1][t - 1]) % mod;
      }
    }
  }
  int N;
  cin >> N;
  for (int g = 0; g < N; g++) {
    int a, b;
    cin >> a >> b;
    if (b > 50) {
      cout << 0 << '\n';
      continue;
    }
    cout << (1LL * fact[b] * dp[a][a][b]) % mod << '\n';
  }
  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;
int dp[1005][1005][51];
int fact[1005];
const int mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(0);
  fact[0] = 1;
  for (int g = 1; g <= 1000; g++) fact[g] = (1LL * g * fact[g - 1]) % mod;
  for (int g = 0; g <= 1000; g++)
    for (int y = 0; y <= 1000; y++) dp[g][y][0] = 1;
  for (int g = 1; g <= 1000; g++) {
    for (int y = 0; y <= 1000; y++) {
      for (int t = 1; t <= 50; t++) {
        if (y == 0) {
          if (t == 1) {
            dp[g][y][t] = (1 + dp[g - 1][y][t]) % mod;
          }
          continue;
        }
        dp[g][y][t] = (dp[g - 1][y][t] + dp[g][y - 1][t] - dp[g - 1][y - 1][t]);
        dp[g][y][t] %= mod;
        dp[g][y][t] += mod;
        dp[g][y][t] %= mod;
        if (g - y > 0)
          dp[g][y][t] = (dp[g][y][t] + dp[g - y - 1][y - 1][t - 1]) % mod;
      }
    }
  }
  int N;
  cin >> N;
  for (int g = 0; g < N; g++) {
    int a, b;
    cin >> a >> b;
    if (b > 50) {
      cout << 0 << '\n';
      continue;
    }
    cout << (1LL * fact[b] * dp[a][a][b]) % mod << '\n';
  }
  return 0;
}

```