#include <bits/stdc++.h>
const long double eps = 1e-9;
const long double pi = acos(-1.0);
using namespace std;
const long long MOD = 1e9 + 7;
const int N = 1e3 + 1;
long long dp[N][55], f[N], c[N][N], ans[N][N];
void inic() {
  f[0] = 1;
  for (int i = 1; i < N; i++) f[i] = (f[i - 1] * 1ll * i) % MOD;
  for (int i = 0; i <= 1000; i++) {
    c[i][0] = c[i][i] = 1;
    for (int j = 1; j < i; j++) {
      c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
      c[i][j] %= MOD;
    }
  }
}
void prec() {
  dp[0][0] = 1;
  for (int new_len = 1; new_len <= 1000; new_len++) {
    for (int last_len = 1000 - new_len; last_len >= 0; last_len--) {
      for (int cnt = 1; cnt <= 50; cnt++) {
        dp[new_len + last_len][cnt] += dp[last_len][cnt - 1];
        dp[new_len + last_len][cnt] %= MOD;
      }
    }
  }
  for (int new_len = 1; new_len <= 1000; new_len++) {
    for (int last_len = 1; last_len <= new_len; last_len++) {
      for (int cnt = 1; cnt <= 50; cnt++) {
        ans[new_len][cnt] += ((dp[last_len][cnt] * f[cnt] % MOD) *
                              c[cnt + new_len - last_len][cnt]) %
                             MOD;
        ans[new_len][cnt] %= MOD;
      }
    }
  }
}
void solve() {
  inic();
  prec();
  int n;
  cin >> n;
  while (n--) {
    int x, y;
    cin >> x >> y;
    cout << ans[x][y] << endl;
  }
}
int main() {
  ios::sync_with_stdio(NULL), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(10);
  int step;
  step = 1;
  for (int i = 1; i <= step; i++) {
    solve();
  }
}

### Prompt

Your task is to create 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>
const long double eps = 1e-9;
const long double pi = acos(-1.0);
using namespace std;
const long long MOD = 1e9 + 7;
const int N = 1e3 + 1;
long long dp[N][55], f[N], c[N][N], ans[N][N];
void inic() {
  f[0] = 1;
  for (int i = 1; i < N; i++) f[i] = (f[i - 1] * 1ll * i) % MOD;
  for (int i = 0; i <= 1000; i++) {
    c[i][0] = c[i][i] = 1;
    for (int j = 1; j < i; j++) {
      c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
      c[i][j] %= MOD;
    }
  }
}
void prec() {
  dp[0][0] = 1;
  for (int new_len = 1; new_len <= 1000; new_len++) {
    for (int last_len = 1000 - new_len; last_len >= 0; last_len--) {
      for (int cnt = 1; cnt <= 50; cnt++) {
        dp[new_len + last_len][cnt] += dp[last_len][cnt - 1];
        dp[new_len + last_len][cnt] %= MOD;
      }
    }
  }
  for (int new_len = 1; new_len <= 1000; new_len++) {
    for (int last_len = 1; last_len <= new_len; last_len++) {
      for (int cnt = 1; cnt <= 50; cnt++) {
        ans[new_len][cnt] += ((dp[last_len][cnt] * f[cnt] % MOD) *
                              c[cnt + new_len - last_len][cnt]) %
                             MOD;
        ans[new_len][cnt] %= MOD;
      }
    }
  }
}
void solve() {
  inic();
  prec();
  int n;
  cin >> n;
  while (n--) {
    int x, y;
    cin >> x >> y;
    cout << ans[x][y] << endl;
  }
}
int main() {
  ios::sync_with_stdio(NULL), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(10);
  int step;
  step = 1;
  for (int i = 1; i <= step; i++) {
    solve();
  }
}

```