#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int MAXN = 1000;
const int MAXM = 50;
int t;
int n, m;
int sum[MAXM + 10][MAXN + 10][MAXN + 10];
int quickpow(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = (long long)res * x % MOD;
    x = (long long)x * x % MOD;
    y >>= 1;
  }
  return res;
}
int ny[MAXN + 10];
void getny() {
  for (int i = 1; i <= MAXN; ++i) ny[i] = quickpow(i, MOD - 2);
}
int main() {
  getny();
  for (int i = 1; i <= MAXM; ++i) {
    for (int j = 1; j <= MAXN; ++j) {
      for (int k = 1; k <= MAXN; ++k) {
        sum[i][j][k] = sum[i][j][k - 1];
        if (i == 1) {
          if (j == k) (++sum[i][j][k]) %= MOD;
        } else if (j >= k)
          (sum[i][j][k] += sum[i - 1][j - k][k - 1]) %= MOD;
      }
    }
  }
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &n, &m);
    if (m > 50) {
      puts("0");
      continue;
    }
    int res = 0, now = 1, a, b;
    for (int i = 2; i <= m; ++i) now = (long long)now * i % MOD;
    for (int i = n; i >= 0; --i) {
      res = ((long long)res + (long long)sum[m][i][i] * now % MOD) % MOD;
      a = n - i + 1 + m;
      b = n - i + 1;
      now = (long long)now * a % MOD * ny[b] % MOD;
    }
    printf("%d\n", res);
  }
  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 = 1e9 + 7;
const int MAXN = 1000;
const int MAXM = 50;
int t;
int n, m;
int sum[MAXM + 10][MAXN + 10][MAXN + 10];
int quickpow(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = (long long)res * x % MOD;
    x = (long long)x * x % MOD;
    y >>= 1;
  }
  return res;
}
int ny[MAXN + 10];
void getny() {
  for (int i = 1; i <= MAXN; ++i) ny[i] = quickpow(i, MOD - 2);
}
int main() {
  getny();
  for (int i = 1; i <= MAXM; ++i) {
    for (int j = 1; j <= MAXN; ++j) {
      for (int k = 1; k <= MAXN; ++k) {
        sum[i][j][k] = sum[i][j][k - 1];
        if (i == 1) {
          if (j == k) (++sum[i][j][k]) %= MOD;
        } else if (j >= k)
          (sum[i][j][k] += sum[i - 1][j - k][k - 1]) %= MOD;
      }
    }
  }
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &n, &m);
    if (m > 50) {
      puts("0");
      continue;
    }
    int res = 0, now = 1, a, b;
    for (int i = 2; i <= m; ++i) now = (long long)now * i % MOD;
    for (int i = n; i >= 0; --i) {
      res = ((long long)res + (long long)sum[m][i][i] * now % MOD) % MOD;
      a = n - i + 1 + m;
      b = n - i + 1;
      now = (long long)now * a % MOD * ny[b] % MOD;
    }
    printf("%d\n", res);
  }
  return 0;
}

```