#include <bits/stdc++.h>
using namespace std;
const int N = 1010, mod = 1000000007, K = 50, T = 33;
int c[N + K + 10][K + N + 10], f[N], dp[N + 10][K + 10][N + 10], q, w, ans, t;
long long sum;
void ent() {
  for (int i = 0; i <= N + K; i++) {
    c[i][i] = 1;
    c[i][0] = 1;
  }
  for (int i = 1; i <= N + K; i++)
    for (int j = 1; j <= K; j++) {
      c[i][j] = ((long long)c[i - 1][j] + c[i - 1][j - 1]) % mod;
      c[i][j] %= mod;
    }
  return;
}
void fa() {
  f[0] = 1;
  for (int i = 1; i <= K; i++) {
    f[i] = ((long long)1 * (1LL * f[i - 1] * i)) % mod;
    f[i] %= mod;
  }
  return;
}
void d() {
  dp[0][0][0] = 1;
  for (int i = 1; i <= N; i++)
    for (int j = 0; j <= K; j++) {
      for (int l = 0; l <= N; l++) {
        dp[i][j][l] = dp[i - 1][j][l];
        if (j > 0 && l >= i)
          dp[i][j][l] = (dp[i - 1][j - 1][l - i] + dp[i - 1][j][l]) % mod;
        dp[i][j][l] %= mod;
      }
    }
  return;
}
int main() {
  ent();
  fa();
  d();
  cin >> t;
  for (int j = 0; j < t; j++) {
    cin >> q >> w;
    sum = 0;
    if (w < K)
      for (int i = 1; i <= q; i++)
        sum = (sum + 1LL * f[w] * c[q - i + w][w] % mod * dp[q][w][i]) % mod %
              mod;
    cout << sum % mod << endl;
  }
}

### Prompt

Your challenge is to write 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 N = 1010, mod = 1000000007, K = 50, T = 33;
int c[N + K + 10][K + N + 10], f[N], dp[N + 10][K + 10][N + 10], q, w, ans, t;
long long sum;
void ent() {
  for (int i = 0; i <= N + K; i++) {
    c[i][i] = 1;
    c[i][0] = 1;
  }
  for (int i = 1; i <= N + K; i++)
    for (int j = 1; j <= K; j++) {
      c[i][j] = ((long long)c[i - 1][j] + c[i - 1][j - 1]) % mod;
      c[i][j] %= mod;
    }
  return;
}
void fa() {
  f[0] = 1;
  for (int i = 1; i <= K; i++) {
    f[i] = ((long long)1 * (1LL * f[i - 1] * i)) % mod;
    f[i] %= mod;
  }
  return;
}
void d() {
  dp[0][0][0] = 1;
  for (int i = 1; i <= N; i++)
    for (int j = 0; j <= K; j++) {
      for (int l = 0; l <= N; l++) {
        dp[i][j][l] = dp[i - 1][j][l];
        if (j > 0 && l >= i)
          dp[i][j][l] = (dp[i - 1][j - 1][l - i] + dp[i - 1][j][l]) % mod;
        dp[i][j][l] %= mod;
      }
    }
  return;
}
int main() {
  ent();
  fa();
  d();
  cin >> t;
  for (int j = 0; j < t; j++) {
    cin >> q >> w;
    sum = 0;
    if (w < K)
      for (int i = 1; i <= q; i++)
        sum = (sum + 1LL * f[w] * c[q - i + w][w] % mod * dp[q][w][i]) % mod %
              mod;
    cout << sum % mod << endl;
  }
}

```