#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1010;
const int MAXV = MAXN << 1;
const int SQRT = 55;
const int MOD = 1E9 + 7;
int mul(int a, int b) { return 1ll * a * b % MOD; }
int add(int a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
  return a;
}
int DP[SQRT][MAXN];
int sum[SQRT][MAXN];
int fact[SQRT];
int choose[MAXV][MAXV];
void precomp() {
  fact[0] = 1;
  for (int i = 1; i < SQRT; ++i) {
    fact[i] = mul(i, fact[i - 1]);
  }
  for (int i = 0; i < MAXV; ++i) {
    choose[i][0] = 1;
    if (i >= 1) {
      for (int k = 1; k <= i; ++k) {
        choose[i][k] = add(choose[i - 1][k], choose[i - 1][k - 1]);
      }
    }
  }
  DP[0][0] = 1;
  for (int i = 1; i < MAXN; ++i) {
    for (int k = SQRT - 1; k >= 1; --k) {
      for (int j = 0; j < MAXN; ++j) {
        if (j >= i) {
          DP[k][j] = add(DP[k][j], DP[k - 1][j - i]);
        }
      }
    }
  }
  for (int i = 1; i < SQRT; ++i) {
    for (int k = 1; k < MAXN; ++k) {
      for (int j = 0; j <= k; ++j) {
        sum[i][k] = add(sum[i][k], mul(DP[i][j], choose[k - j + i][i]));
      }
    }
  }
}
int main() {
  precomp();
  int T;
  scanf("%d", &T);
  for (int i = 0; i < T; ++i) {
    int N, K;
    scanf("%d %d", &N, &K);
    if (K >= SQRT) {
      printf("0\n");
      continue;
    }
    int res = mul(sum[K][N], fact[K]);
    printf("%d\n", res);
  }
  return 0;
}

### 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;
const int MAXN = 1010;
const int MAXV = MAXN << 1;
const int SQRT = 55;
const int MOD = 1E9 + 7;
int mul(int a, int b) { return 1ll * a * b % MOD; }
int add(int a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
  return a;
}
int DP[SQRT][MAXN];
int sum[SQRT][MAXN];
int fact[SQRT];
int choose[MAXV][MAXV];
void precomp() {
  fact[0] = 1;
  for (int i = 1; i < SQRT; ++i) {
    fact[i] = mul(i, fact[i - 1]);
  }
  for (int i = 0; i < MAXV; ++i) {
    choose[i][0] = 1;
    if (i >= 1) {
      for (int k = 1; k <= i; ++k) {
        choose[i][k] = add(choose[i - 1][k], choose[i - 1][k - 1]);
      }
    }
  }
  DP[0][0] = 1;
  for (int i = 1; i < MAXN; ++i) {
    for (int k = SQRT - 1; k >= 1; --k) {
      for (int j = 0; j < MAXN; ++j) {
        if (j >= i) {
          DP[k][j] = add(DP[k][j], DP[k - 1][j - i]);
        }
      }
    }
  }
  for (int i = 1; i < SQRT; ++i) {
    for (int k = 1; k < MAXN; ++k) {
      for (int j = 0; j <= k; ++j) {
        sum[i][k] = add(sum[i][k], mul(DP[i][j], choose[k - j + i][i]));
      }
    }
  }
}
int main() {
  precomp();
  int T;
  scanf("%d", &T);
  for (int i = 0; i < T; ++i) {
    int N, K;
    scanf("%d %d", &N, &K);
    if (K >= SQRT) {
      printf("0\n");
      continue;
    }
    int res = mul(sum[K][N], fact[K]);
    printf("%d\n", res);
  }
  return 0;
}

```