#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long dp[55][1010];
long long v[55][1010];
long long ke[55];
long long bin[2020][2020];
void cv(int k, int n) {
  int rs = n - k;
  if (rs < 0) return;
  for (int i = 0; i <= rs; i++) {
    v[k][n] += bin[k + rs - i][k] * dp[k][i];
    v[k][n] %= mod;
  }
  v[k][n] *= ke[k];
  v[k][n] %= mod;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  bin[0][0] = 1;
  for (int i = 1; i <= 2000; i++) {
    bin[i][0] = 1;
    for (int j = 1; j <= 2000; j++) {
      bin[i][j] = bin[i - 1][j - 1] + bin[i - 1][j];
      bin[i][j] %= mod;
    }
  }
  dp[0][0] = 1;
  for (int t = 0; t <= 1000; t++) {
    for (int j = 50; j > 0; j--) {
      for (int i = 1000; i - t >= 0; i--) {
        dp[j][i] += dp[j - 1][i - t];
        dp[j][i] %= mod;
      }
    }
  }
  ke[0] = 1;
  for (long long i = 1; i <= 50; i++) {
    ke[i] = ke[i - 1] * i;
    ke[i] %= mod;
  }
  for (int i = 1; i <= 50; i++) {
    for (int j = 1; j <= 1000; j++) {
      cv(i, j);
    }
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int n, k;
    cin >> n >> k;
    if (k > 50)
      cout << 0 << '\n';
    else
      cout << v[k][n] << '\n';
  }
}

### 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;
const long long mod = 1e9 + 7;
long long dp[55][1010];
long long v[55][1010];
long long ke[55];
long long bin[2020][2020];
void cv(int k, int n) {
  int rs = n - k;
  if (rs < 0) return;
  for (int i = 0; i <= rs; i++) {
    v[k][n] += bin[k + rs - i][k] * dp[k][i];
    v[k][n] %= mod;
  }
  v[k][n] *= ke[k];
  v[k][n] %= mod;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  bin[0][0] = 1;
  for (int i = 1; i <= 2000; i++) {
    bin[i][0] = 1;
    for (int j = 1; j <= 2000; j++) {
      bin[i][j] = bin[i - 1][j - 1] + bin[i - 1][j];
      bin[i][j] %= mod;
    }
  }
  dp[0][0] = 1;
  for (int t = 0; t <= 1000; t++) {
    for (int j = 50; j > 0; j--) {
      for (int i = 1000; i - t >= 0; i--) {
        dp[j][i] += dp[j - 1][i - t];
        dp[j][i] %= mod;
      }
    }
  }
  ke[0] = 1;
  for (long long i = 1; i <= 50; i++) {
    ke[i] = ke[i - 1] * i;
    ke[i] %= mod;
  }
  for (int i = 1; i <= 50; i++) {
    for (int j = 1; j <= 1000; j++) {
      cv(i, j);
    }
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int n, k;
    cin >> n >> k;
    if (k > 50)
      cout << 0 << '\n';
    else
      cout << v[k][n] << '\n';
  }
}

```