#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1123;
const int MAXM = 54;
const int P = 1000000007;
int f[2][MAXM][MAXN];
int ans[MAXN][MAXN];
int fac[MAXN];
int C[MAXN][MAXN];
void add(int &a, int b) { a = (a + b) % P; }
int main() {
  int n, m;
  int i, j, k;
  n = 1000;
  m = 50;
  f[0][0][0] = 1;
  for (i = 0; i < n; i++) {
    memset(f[(i + 1) & 1], 0, sizeof f[0]);
    for (j = 0; j <= m; j++)
      for (k = 0; k <= n; k++) {
        add(f[(i + 1) & 1][j][k], f[i & 1][j][k]);
        if (j + 1 <= m && k + i <= n)
          add(f[(i + 1) & 1][j + 1][k + i], f[i & 1][j][k]);
      }
  }
  fac[0] = 1;
  for (i = 1; i <= n; i++) fac[i] = (long long)fac[i - 1] * i % P;
  for (i = 0; i <= n; i++) {
    C[i][0] = 1;
    for (j = 1; j <= i; j++) {
      add(C[i][j], C[i - 1][j - 1]);
      add(C[i][j], C[i - 1][j]);
    }
  }
  for (i = 0; i <= n; i++)
    for (j = 0; j <= m; j++)
      for (k = 0; k <= i; k++) {
        int t = (long long)fac[j] * f[0][j][k] % P * C[i - k][j] % P;
        add(ans[i][j], t);
      }
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d", &n, &m);
    printf("%d\n", ans[n][m]);
  }
  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;
const int MAXN = 1123;
const int MAXM = 54;
const int P = 1000000007;
int f[2][MAXM][MAXN];
int ans[MAXN][MAXN];
int fac[MAXN];
int C[MAXN][MAXN];
void add(int &a, int b) { a = (a + b) % P; }
int main() {
  int n, m;
  int i, j, k;
  n = 1000;
  m = 50;
  f[0][0][0] = 1;
  for (i = 0; i < n; i++) {
    memset(f[(i + 1) & 1], 0, sizeof f[0]);
    for (j = 0; j <= m; j++)
      for (k = 0; k <= n; k++) {
        add(f[(i + 1) & 1][j][k], f[i & 1][j][k]);
        if (j + 1 <= m && k + i <= n)
          add(f[(i + 1) & 1][j + 1][k + i], f[i & 1][j][k]);
      }
  }
  fac[0] = 1;
  for (i = 1; i <= n; i++) fac[i] = (long long)fac[i - 1] * i % P;
  for (i = 0; i <= n; i++) {
    C[i][0] = 1;
    for (j = 1; j <= i; j++) {
      add(C[i][j], C[i - 1][j - 1]);
      add(C[i][j], C[i - 1][j]);
    }
  }
  for (i = 0; i <= n; i++)
    for (j = 0; j <= m; j++)
      for (k = 0; k <= i; k++) {
        int t = (long long)fac[j] * f[0][j][k] % P * C[i - k][j] % P;
        add(ans[i][j], t);
      }
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d", &n, &m);
    printf("%d\n", ans[n][m]);
  }
  return 0;
}

```