#include <bits/stdc++.h>
using namespace std;
int n, k;
int M[47][1001][1001];
int DP(int id, int ac, int l) {
  if (id == 1) return ac >= l;
  if (l > ac) return 0;
  int &ans = M[id][ac][l];
  if (ans >= 0) return ans;
  ans = DP(id - 1, ac - l, l + 1) + DP(id, ac, l + 1);
  if (ans >= 1000000007) ans -= 1000000007;
  return ans;
}
long long F[1001];
int C[1001][1001];
void pre() {
  memset(M, -1, sizeof M);
  F[0] = 1;
  C[0][0] = 1;
  for (int i = 1; i < 1001; ++i) {
    F[i] = F[i - 1] * i;
    if (F[i] >= 1000000007) F[i] %= 1000000007;
    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];
      if (C[i][j] >= 1000000007) C[i][j] -= 1000000007;
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  pre();
  int tc;
  cin >> tc;
  while (tc--) {
    cin >> n >> k;
    long long ans = 0, t;
    for (int i = (k - 1) * k / 2; i <= n - k; ++i) {
      t = (long long)DP(k, i, 0) * C[n - i][k];
      if (t >= 1000000007) t %= 1000000007;
      ans += t;
      if (ans >= 1000000007) ans -= 1000000007;
    }
    ans *= F[k];
    if (ans >= 1000000007) ans %= 1000000007;
    cout << ans << endl;
  }
}

### Prompt

Please formulate 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;
int n, k;
int M[47][1001][1001];
int DP(int id, int ac, int l) {
  if (id == 1) return ac >= l;
  if (l > ac) return 0;
  int &ans = M[id][ac][l];
  if (ans >= 0) return ans;
  ans = DP(id - 1, ac - l, l + 1) + DP(id, ac, l + 1);
  if (ans >= 1000000007) ans -= 1000000007;
  return ans;
}
long long F[1001];
int C[1001][1001];
void pre() {
  memset(M, -1, sizeof M);
  F[0] = 1;
  C[0][0] = 1;
  for (int i = 1; i < 1001; ++i) {
    F[i] = F[i - 1] * i;
    if (F[i] >= 1000000007) F[i] %= 1000000007;
    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];
      if (C[i][j] >= 1000000007) C[i][j] -= 1000000007;
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  pre();
  int tc;
  cin >> tc;
  while (tc--) {
    cin >> n >> k;
    long long ans = 0, t;
    for (int i = (k - 1) * k / 2; i <= n - k; ++i) {
      t = (long long)DP(k, i, 0) * C[n - i][k];
      if (t >= 1000000007) t %= 1000000007;
      ans += t;
      if (ans >= 1000000007) ans -= 1000000007;
    }
    ans *= F[k];
    if (ans >= 1000000007) ans %= 1000000007;
    cout << ans << endl;
  }
}

```