#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
const int N = 1000;
int sum[45][N + 1][N + 1];
int ans[45][N + 1];
int dp[2][N + 1][N + 1];
int A[N + 1][N + 1];
int fix(long long a, long long b) { return (a + b) % mod; }
void init() {
  for (int i = 1; i <= N; i++) {
    A[i][0] = 1;
    A[i][1] = i;
    for (int j = 2; j <= N; j++) {
      A[i][j] = A[i][j - 1] * 1ll * (i - j + 1) % mod;
    }
  }
  dp[0][0][0] = 1;
  sum[0][0][0] = 1;
  for (int j = 0; j <= N; j++) {
    for (int k = 1; k <= N; k++) {
      sum[0][j][k] = sum[0][j][k - 1] + dp[0][j][k];
    }
  }
  for (int i = 1; i <= 44; i++) {
    memset(dp[i & 1], 0, sizeof(dp[i & 1]));
    for (int j = i; j <= N; j++) {
      for (int k = j; k <= N; k++) {
        dp[i & 1][k][j] = fix(sum[i - 1][k - j][j - 1], dp[i & 1][k][j]);
      }
    }
    for (int k = 1; k <= N; k++) {
      for (int j = 1; j <= N; j++) {
        sum[i][k][j] = fix(sum[i][k][j - 1], dp[i & 1][k][j]);
      }
    }
  }
  for (int i = 1; i <= 44; i++) {
    int nmin = i * (i + 1) / 2;
    for (int n = nmin; n <= N; n++) {
      for (int j = nmin; j <= n; j++) {
        int spaceNum = n - j;
        int f = sum[i][j][N];
        int h = A[spaceNum + i][i];
        ans[i][n] = fix(ans[i][n], h * 1ll * f);
      }
    }
  }
}
int main() {
  int T;
  init();
  scanf("%d", &T);
  while (T--) {
    int a, b;
    scanf("%d%d", &a, &b);
    if (b > 44)
      printf("0\n");
    else
      printf("%d\n", ans[b][a]);
  }
  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 = 1000000007;
const int N = 1000;
int sum[45][N + 1][N + 1];
int ans[45][N + 1];
int dp[2][N + 1][N + 1];
int A[N + 1][N + 1];
int fix(long long a, long long b) { return (a + b) % mod; }
void init() {
  for (int i = 1; i <= N; i++) {
    A[i][0] = 1;
    A[i][1] = i;
    for (int j = 2; j <= N; j++) {
      A[i][j] = A[i][j - 1] * 1ll * (i - j + 1) % mod;
    }
  }
  dp[0][0][0] = 1;
  sum[0][0][0] = 1;
  for (int j = 0; j <= N; j++) {
    for (int k = 1; k <= N; k++) {
      sum[0][j][k] = sum[0][j][k - 1] + dp[0][j][k];
    }
  }
  for (int i = 1; i <= 44; i++) {
    memset(dp[i & 1], 0, sizeof(dp[i & 1]));
    for (int j = i; j <= N; j++) {
      for (int k = j; k <= N; k++) {
        dp[i & 1][k][j] = fix(sum[i - 1][k - j][j - 1], dp[i & 1][k][j]);
      }
    }
    for (int k = 1; k <= N; k++) {
      for (int j = 1; j <= N; j++) {
        sum[i][k][j] = fix(sum[i][k][j - 1], dp[i & 1][k][j]);
      }
    }
  }
  for (int i = 1; i <= 44; i++) {
    int nmin = i * (i + 1) / 2;
    for (int n = nmin; n <= N; n++) {
      for (int j = nmin; j <= n; j++) {
        int spaceNum = n - j;
        int f = sum[i][j][N];
        int h = A[spaceNum + i][i];
        ans[i][n] = fix(ans[i][n], h * 1ll * f);
      }
    }
  }
}
int main() {
  int T;
  init();
  scanf("%d", &T);
  while (T--) {
    int a, b;
    scanf("%d%d", &a, &b);
    if (b > 44)
      printf("0\n");
    else
      printf("%d\n", ans[b][a]);
  }
  return 0;
}

```