#include <bits/stdc++.h>
using namespace std;
long long fkt[2005];
long long com[2005][2005];
void init() {
  long long i, j;
  com[0][0] = 1;
  fkt[0] = 1;
  for (i = 1; i <= 2000; i++) {
    fkt[i] = (fkt[i - 1] * i) % 1000000007;
    com[i][0] = 1;
    for (j = 1; j <= i; j++)
      com[i][j] = (com[i - 1][j] + com[i - 1][j - 1]) % 1000000007;
  }
}
int dp[1005][55][1005];
int solve(int len, int cnt, int mx) {
  if (!cnt) return !len;
  if (len < 0 || cnt < 0 || !mx) return 0;
  if (dp[len][cnt][mx] != -1) return dp[len][cnt][mx];
  return dp[len][cnt][mx] =
             (solve(len - mx, cnt - 1, mx - 1) + solve(len, cnt, mx - 1)) %
             1000000007;
}
void Solve_but_really_solve_this_time() {
  int i, j, n, a, b, res = 0;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> a >> b;
    res = 0;
    if (b > 50) {
      cout << 0 << endl;
      continue;
    }
    for (j = 1; j <= a; j++)
      res = (res +
             (((solve(j, b, j) * fkt[b]) % 1000000007) * com[a - j + b][b]) %
                 1000000007) %
            1000000007;
    cout << res << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  memset(dp, -1, sizeof dp);
  init();
  Solve_but_really_solve_this_time();
  return 0;
}

### 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 fkt[2005];
long long com[2005][2005];
void init() {
  long long i, j;
  com[0][0] = 1;
  fkt[0] = 1;
  for (i = 1; i <= 2000; i++) {
    fkt[i] = (fkt[i - 1] * i) % 1000000007;
    com[i][0] = 1;
    for (j = 1; j <= i; j++)
      com[i][j] = (com[i - 1][j] + com[i - 1][j - 1]) % 1000000007;
  }
}
int dp[1005][55][1005];
int solve(int len, int cnt, int mx) {
  if (!cnt) return !len;
  if (len < 0 || cnt < 0 || !mx) return 0;
  if (dp[len][cnt][mx] != -1) return dp[len][cnt][mx];
  return dp[len][cnt][mx] =
             (solve(len - mx, cnt - 1, mx - 1) + solve(len, cnt, mx - 1)) %
             1000000007;
}
void Solve_but_really_solve_this_time() {
  int i, j, n, a, b, res = 0;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> a >> b;
    res = 0;
    if (b > 50) {
      cout << 0 << endl;
      continue;
    }
    for (j = 1; j <= a; j++)
      res = (res +
             (((solve(j, b, j) * fkt[b]) % 1000000007) * com[a - j + b][b]) %
                 1000000007) %
            1000000007;
    cout << res << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  memset(dp, -1, sizeof dp);
  init();
  Solve_but_really_solve_this_time();
  return 0;
}

```