#include <bits/stdc++.h>
using namespace std;
long long dp[1005][1005];
long long dpend[1005][1005];
long long ans[1005][1005];
long long c[1005][1005];
long long sum[1005][1005];
long long f[1005];
int main() {
  srand((unsigned int)time(NULL));
  int t;
  cin >> t;
  for (int i = 0; i <= 1000; i++) {
    for (int j = 0; j <= i; j++) {
      if (j == 0 || i == j)
        c[i][j] = 1LL;
      else
        c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % 1000000007;
    }
  }
  f[0] = 1LL;
  for (int i = 1; i <= 1000; i++) f[i] = (f[i - 1] * i) % 1000000007;
  dp[0][0] = 1;
  for (int i = 1; i <= 1000; i++) {
    for (int sum = 0; sum <= 1000; sum++) {
      if (!dp[sum][i - 1]) continue;
      int tot = 0;
      for (int k = sum; k <= 1000; k += i) {
        dpend[k][i] = (dpend[k][i] + dp[sum][i - 1]) % 1000000007;
        if (k != sum) dp[k][i] = (dp[k][i] + dp[sum][i - 1]) % 1000000007;
      }
    }
  }
  for (int i = 0; i <= 1000; i++)
    for (int j = 1; j <= 1000; j++)
      dpend[i][j] = (dpend[i][j] * f[j]) % 1000000007;
  for (int k = 2; k <= 1000; k++) {
    sum[k][0] = dpend[0][k];
    long long tot = 0;
    for (int xy = 1; xy <= 1000; xy++) {
      tot = (tot + dpend[xy - 1][k]) % 1000000007;
      sum[k][xy] = sum[k][xy - 1];
      sum[k][xy] += tot;
      sum[k][xy] = (sum[k][xy] + dpend[xy][k]) % 1000000007;
    }
  }
  while (t--) {
    int n, k;
    scanf("%d %d", &n, &k);
    if (k == 1) {
      printf("%d\n", n * (n + 1) / 2);
    } else {
      long long ret = 0LL;
      for (int i = 0; i <= n - k; i++) {
        ret =
            (ret + (sum[k][i] * c[n - i - 2][k - 2]) % 1000000007) % 1000000007;
      }
      printf("%lld\n", ret);
    }
  }
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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 dp[1005][1005];
long long dpend[1005][1005];
long long ans[1005][1005];
long long c[1005][1005];
long long sum[1005][1005];
long long f[1005];
int main() {
  srand((unsigned int)time(NULL));
  int t;
  cin >> t;
  for (int i = 0; i <= 1000; i++) {
    for (int j = 0; j <= i; j++) {
      if (j == 0 || i == j)
        c[i][j] = 1LL;
      else
        c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % 1000000007;
    }
  }
  f[0] = 1LL;
  for (int i = 1; i <= 1000; i++) f[i] = (f[i - 1] * i) % 1000000007;
  dp[0][0] = 1;
  for (int i = 1; i <= 1000; i++) {
    for (int sum = 0; sum <= 1000; sum++) {
      if (!dp[sum][i - 1]) continue;
      int tot = 0;
      for (int k = sum; k <= 1000; k += i) {
        dpend[k][i] = (dpend[k][i] + dp[sum][i - 1]) % 1000000007;
        if (k != sum) dp[k][i] = (dp[k][i] + dp[sum][i - 1]) % 1000000007;
      }
    }
  }
  for (int i = 0; i <= 1000; i++)
    for (int j = 1; j <= 1000; j++)
      dpend[i][j] = (dpend[i][j] * f[j]) % 1000000007;
  for (int k = 2; k <= 1000; k++) {
    sum[k][0] = dpend[0][k];
    long long tot = 0;
    for (int xy = 1; xy <= 1000; xy++) {
      tot = (tot + dpend[xy - 1][k]) % 1000000007;
      sum[k][xy] = sum[k][xy - 1];
      sum[k][xy] += tot;
      sum[k][xy] = (sum[k][xy] + dpend[xy][k]) % 1000000007;
    }
  }
  while (t--) {
    int n, k;
    scanf("%d %d", &n, &k);
    if (k == 1) {
      printf("%d\n", n * (n + 1) / 2);
    } else {
      long long ret = 0LL;
      for (int i = 0; i <= n - k; i++) {
        ret =
            (ret + (sum[k][i] * c[n - i - 2][k - 2]) % 1000000007) % 1000000007;
      }
      printf("%lld\n", ret);
    }
  }
}

```