#include <bits/stdc++.h>
using namespace std;
int nCr[2 * 1006][2 * 1006], fact[2 * 1006];
void pre() {
  int i, j;
  fact[0] = nCr[0][0] = 1;
  for (i = 1; i < 2 * 1006; ++i) {
    fact[i] = ((long long)fact[i - 1] * i) % 1000000007;
    nCr[i][1] = i, nCr[i][0] = nCr[i][i] = 1;
  }
  for (i = 2; i < 2 * 1006; ++i) {
    for (j = 2; j < i; ++j)
      nCr[i][j] = (nCr[i - 1][j] + nCr[i - 1][j - 1]) % 1000000007;
  }
}
int dp[1006][50][1006];
int rec(int n, int k, int d) {
  if (n < 0) return 0;
  if (k == 0) return 1;
  if (d >= n) return 0;
  int &ret = dp[n][k][d];
  if (ret != -1) return ret;
  ret = rec(n - d - 1, k - 1, d + 1);
  if (d) ret = (ret + rec(n - 1, k, d + 1)) % 1000000007;
  return ret % 1000000007;
}
int main() {
  pre();
  memset(dp, -1, sizeof(dp));
  int T, N, K, n, ret;
  scanf("%d", &T);
  while (T--) {
    scanf("%d %d", &N, &K);
    if (K >= 45) {
      puts("0");
      continue;
    }
    ret = 0;
    for (n = 1; n <= N; ++n) {
      ret += ((((long long)rec(n, K, 0) * fact[K]) % 1000000007) *
              (long long)nCr[N - n + K][K]) %
             1000000007;
      ret %= 1000000007;
    }
    printf("%d\n", ret);
  }
  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;
int nCr[2 * 1006][2 * 1006], fact[2 * 1006];
void pre() {
  int i, j;
  fact[0] = nCr[0][0] = 1;
  for (i = 1; i < 2 * 1006; ++i) {
    fact[i] = ((long long)fact[i - 1] * i) % 1000000007;
    nCr[i][1] = i, nCr[i][0] = nCr[i][i] = 1;
  }
  for (i = 2; i < 2 * 1006; ++i) {
    for (j = 2; j < i; ++j)
      nCr[i][j] = (nCr[i - 1][j] + nCr[i - 1][j - 1]) % 1000000007;
  }
}
int dp[1006][50][1006];
int rec(int n, int k, int d) {
  if (n < 0) return 0;
  if (k == 0) return 1;
  if (d >= n) return 0;
  int &ret = dp[n][k][d];
  if (ret != -1) return ret;
  ret = rec(n - d - 1, k - 1, d + 1);
  if (d) ret = (ret + rec(n - 1, k, d + 1)) % 1000000007;
  return ret % 1000000007;
}
int main() {
  pre();
  memset(dp, -1, sizeof(dp));
  int T, N, K, n, ret;
  scanf("%d", &T);
  while (T--) {
    scanf("%d %d", &N, &K);
    if (K >= 45) {
      puts("0");
      continue;
    }
    ret = 0;
    for (n = 1; n <= N; ++n) {
      ret += ((((long long)rec(n, K, 0) * fact[K]) % 1000000007) *
              (long long)nCr[N - n + K][K]) %
             1000000007;
      ret %= 1000000007;
    }
    printf("%d\n", ret);
  }
  return 0;
}

```