#include <bits/stdc++.h>
using namespace std;
const int MOD = (int)1e9 + 7;
const int MAXN = 1005;
long long f[MAXN][63], jie[63], c[MAXN][63];
int ans[MAXN][63];
void init() {
  memset(ans, -1, sizeof(ans));
  for (int i = 1; i < MAXN; i++) f[i][1] = 1;
  for (int i = 1; i < MAXN; i++) {
    for (int j = 2; j < 63 && j <= i; j++) {
      f[i][j] = (f[i - j][j - 1] + f[i - j][j]) % MOD;
    }
  }
  c[0][0] = 1;
  for (int i = 1; i < MAXN; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i && j < 63; j++) {
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % MOD;
    }
  }
  jie[0] = 1;
  for (int i = 1; i < 63; i++) jie[i] = jie[i - 1] * i % MOD;
}
int solve(int n, int k) {
  if (ans[n][k] != -1) return ans[n][k];
  long long res = 0;
  int b = k * (k + 1) / 2;
  for (int i = b; i <= n; i++) {
    (res += f[i][k] * jie[k] % MOD * c[n - i + k][k] % MOD) %= MOD;
  }
  ans[n][k] = (int)res;
  return ans[n][k];
}
int main() {
  init();
  int t;
  scanf("%d", &t);
  for (int i = 1, n, k; i <= t; i++) {
    scanf("%d %d", &n, &k);
    if (k * (k + 1) > 2 * n)
      puts("0");
    else
      printf("%d\n", solve(n, k));
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 MOD = (int)1e9 + 7;
const int MAXN = 1005;
long long f[MAXN][63], jie[63], c[MAXN][63];
int ans[MAXN][63];
void init() {
  memset(ans, -1, sizeof(ans));
  for (int i = 1; i < MAXN; i++) f[i][1] = 1;
  for (int i = 1; i < MAXN; i++) {
    for (int j = 2; j < 63 && j <= i; j++) {
      f[i][j] = (f[i - j][j - 1] + f[i - j][j]) % MOD;
    }
  }
  c[0][0] = 1;
  for (int i = 1; i < MAXN; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i && j < 63; j++) {
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % MOD;
    }
  }
  jie[0] = 1;
  for (int i = 1; i < 63; i++) jie[i] = jie[i - 1] * i % MOD;
}
int solve(int n, int k) {
  if (ans[n][k] != -1) return ans[n][k];
  long long res = 0;
  int b = k * (k + 1) / 2;
  for (int i = b; i <= n; i++) {
    (res += f[i][k] * jie[k] % MOD * c[n - i + k][k] % MOD) %= MOD;
  }
  ans[n][k] = (int)res;
  return ans[n][k];
}
int main() {
  init();
  int t;
  scanf("%d", &t);
  for (int i = 1, n, k; i <= t; i++) {
    scanf("%d %d", &n, &k);
    if (k * (k + 1) > 2 * n)
      puts("0");
    else
      printf("%d\n", solve(n, k));
  }
  return 0;
}

```