#include <bits/stdc++.h>
using namespace std;
int name[1001][1001][46];
bool vis[1001][1001][46];
int dp_func(int n, int pos, int k) {
  if (k == 0) return 1;
  if (n == 0) return 0;
  if (pos <= 0) return 0;
  if (vis[n][pos][k]) return name[n][pos][k];
  vis[n][pos][k] = 1;
  long long ret = 0;
  if (n >= pos) {
    ret += dp_func(n - pos, pos - 1, k - 1);
  }
  ret += dp_func(n - 1, pos, k);
  ret += dp_func(n, pos - 1, k);
  ret -= dp_func(n - 1, pos - 1, k);
  if (ret < 0) ret += 1000000007;
  ret %= 1000000007;
  return name[n][pos][k] = ret;
}
long long fact[2000];
int main() {
  fact[0] = 1;
  for (int i = 1; i <= 1000; i++) {
    fact[i] = (fact[i - 1] * i) % 1000000007;
  }
  int i, j, k, l, m, n, ts;
  scanf("%d", &ts);
  while (ts--) {
    scanf("%d%d", &n, &k);
    if (k > 45) {
      printf("0\n");
      continue;
    }
    long long ans = dp_func(n, n, k);
    ans *= fact[k];
    ans %= 1000000007;
    printf("%lld\n", ans);
  }
}

### Prompt

Please provide a CPP coded solution 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;
int name[1001][1001][46];
bool vis[1001][1001][46];
int dp_func(int n, int pos, int k) {
  if (k == 0) return 1;
  if (n == 0) return 0;
  if (pos <= 0) return 0;
  if (vis[n][pos][k]) return name[n][pos][k];
  vis[n][pos][k] = 1;
  long long ret = 0;
  if (n >= pos) {
    ret += dp_func(n - pos, pos - 1, k - 1);
  }
  ret += dp_func(n - 1, pos, k);
  ret += dp_func(n, pos - 1, k);
  ret -= dp_func(n - 1, pos - 1, k);
  if (ret < 0) ret += 1000000007;
  ret %= 1000000007;
  return name[n][pos][k] = ret;
}
long long fact[2000];
int main() {
  fact[0] = 1;
  for (int i = 1; i <= 1000; i++) {
    fact[i] = (fact[i - 1] * i) % 1000000007;
  }
  int i, j, k, l, m, n, ts;
  scanf("%d", &ts);
  while (ts--) {
    scanf("%d%d", &n, &k);
    if (k > 45) {
      printf("0\n");
      continue;
    }
    long long ans = dp_func(n, n, k);
    ans *= fact[k];
    ans %= 1000000007;
    printf("%lld\n", ans);
  }
}

```