#include <bits/stdc++.h>
using namespace std;
long long C[2 * 1005][2 * 1005], dp[1005][1005], cnt[1005][1005], A[1005];
void Init() {
  for (int i = 0; i < 1005 * 2; i++) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; j++)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % 1000000007;
  }
  dp[0][0] = 1;
  for (int i = 1; i < 1005; i++) {
    int now = i * (i + 1) / 2;
    if (now > 1005) break;
    for (int j = now; j < 1005; j++)
      dp[i][j] = (dp[i][j - i] + dp[i - 1][j - i]) % 1000000007;
  }
  A[0] = 1;
  for (int i = 1; i < 1005; i++) A[i] = A[i - 1] * i % 1000000007;
  for (int i = 1; i < 1005; i++) {
    int now = i * (i + 1) / 2;
    if (now > 1005) break;
    for (int j = now; j < 1005; j++) {
      for (int k = j; k < 1005; k++) {
        if (k == j)
          cnt[k][i] = (cnt[k][i] + dp[i][j]) % 1000000007;
        else
          cnt[k][i] = (cnt[k][i] +
                       dp[i][j] * C[i + 1 + k - j - 1][k - j] % 1000000007) %
                      1000000007;
      }
    }
  }
  for (int i = 0; i < 1005; i++)
    for (int j = 0; j < 1005; j++) cnt[i][j] = cnt[i][j] * A[j] % 1000000007;
}
int main() {
  Init();
  int q;
  scanf("%d", &q);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    printf("%lld\n", cnt[x][y]);
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 C[2 * 1005][2 * 1005], dp[1005][1005], cnt[1005][1005], A[1005];
void Init() {
  for (int i = 0; i < 1005 * 2; i++) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; j++)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % 1000000007;
  }
  dp[0][0] = 1;
  for (int i = 1; i < 1005; i++) {
    int now = i * (i + 1) / 2;
    if (now > 1005) break;
    for (int j = now; j < 1005; j++)
      dp[i][j] = (dp[i][j - i] + dp[i - 1][j - i]) % 1000000007;
  }
  A[0] = 1;
  for (int i = 1; i < 1005; i++) A[i] = A[i - 1] * i % 1000000007;
  for (int i = 1; i < 1005; i++) {
    int now = i * (i + 1) / 2;
    if (now > 1005) break;
    for (int j = now; j < 1005; j++) {
      for (int k = j; k < 1005; k++) {
        if (k == j)
          cnt[k][i] = (cnt[k][i] + dp[i][j]) % 1000000007;
        else
          cnt[k][i] = (cnt[k][i] +
                       dp[i][j] * C[i + 1 + k - j - 1][k - j] % 1000000007) %
                      1000000007;
      }
    }
  }
  for (int i = 0; i < 1005; i++)
    for (int j = 0; j < 1005; j++) cnt[i][j] = cnt[i][j] * A[j] % 1000000007;
}
int main() {
  Init();
  int q;
  scanf("%d", &q);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    printf("%lld\n", cnt[x][y]);
  }
  return 0;
}

```