#include <bits/stdc++.h>
using namespace std;
const long long P = 1000000007;
const int N = 1000;
int t;
long long dp[1000 + 10][50];
long long ans[1000 + 10][50];
long long C[1050 + 10][1000 + 10];
long long jc[1000 + 10];
void pre() {
  dp[0][0] = 1;
  for (int i = 1; i <= N; i++) {
    for (int j = N; j >= i; j--) {
      for (int k = 50; k >= 1; k--) {
        dp[j][k] = (dp[j][k] + dp[j - i][k - 1]) % P;
      }
    }
  }
  jc[0] = 1;
  for (int i = 1; i <= 50; i++) jc[i] = (long long)jc[i - 1] * i % P;
  for (int i = 0; i <= N; i++) C[i][0] = 1;
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= i; j++) {
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % P;
    }
  }
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= 50; j++) {
      dp[i][j] = ((long long)dp[i][j] * jc[j]) % P;
    }
  }
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= 50; j++) {
      for (int k = 1; k <= i; k++) {
        ans[i][j] = (ans[i][j] + (long long)dp[k][j] * C[i - k + j][j]) % P;
      }
    }
  }
}
int main() {
  pre();
  scanf("%d", &t);
  while (t--) {
    int n, k;
    scanf("%d%d", &n, &k);
    if ((long long)(1 + k) * k / 2 > n)
      printf("0\n");
    else
      printf("%I64d\n", ans[n][k]);
  }
  return 0;
}

### Prompt

Generate 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;
const long long P = 1000000007;
const int N = 1000;
int t;
long long dp[1000 + 10][50];
long long ans[1000 + 10][50];
long long C[1050 + 10][1000 + 10];
long long jc[1000 + 10];
void pre() {
  dp[0][0] = 1;
  for (int i = 1; i <= N; i++) {
    for (int j = N; j >= i; j--) {
      for (int k = 50; k >= 1; k--) {
        dp[j][k] = (dp[j][k] + dp[j - i][k - 1]) % P;
      }
    }
  }
  jc[0] = 1;
  for (int i = 1; i <= 50; i++) jc[i] = (long long)jc[i - 1] * i % P;
  for (int i = 0; i <= N; i++) C[i][0] = 1;
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= i; j++) {
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % P;
    }
  }
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= 50; j++) {
      dp[i][j] = ((long long)dp[i][j] * jc[j]) % P;
    }
  }
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= 50; j++) {
      for (int k = 1; k <= i; k++) {
        ans[i][j] = (ans[i][j] + (long long)dp[k][j] * C[i - k + j][j]) % P;
      }
    }
  }
}
int main() {
  pre();
  scanf("%d", &t);
  while (t--) {
    int n, k;
    scanf("%d%d", &n, &k);
    if ((long long)(1 + k) * k / 2 > n)
      printf("0\n");
    else
      printf("%I64d\n", ans[n][k]);
  }
  return 0;
}

```