#include <bits/stdc++.h>
using namespace std;
const int T = 2 * 1e5 + 10, N = 1010, mod = 1e9 + 7, K = 50;
int dp[N][N][K], c[N][N];
long long fact[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  c[0][0] = 1;
  for (int i = 1; i < N; i++) {
    c[0][i] = 1;
    c[i][i] = 1;
    for (int j = 1; j < i; j++) {
      c[j][i] = c[j - 1][i - 1] + c[j][i - 1];
      c[j][i] %= mod;
    }
  }
  fact[0] = 1;
  for (int i = 1; i < N; i++) {
    fact[i] = fact[i - 1] * i;
    fact[i] %= mod;
  }
  dp[0][0][0] = 1;
  for (int i = 1; i < N; i++) dp[i][0][0] = 1;
  for (int i = 1; i < N; i++) {
    for (int j = 0; j < N; j++) {
      for (int k = 1; k < K; k++) {
        dp[i][j][k] = dp[i - 1][j][k];
        if (j >= i) dp[i][j][k] += dp[i - 1][j - i][k - 1];
        dp[i][j][k] %= mod;
      }
    }
  }
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    long long n, k;
    cin >> n >> k;
    if (k >= 50) {
      cout << 0 << '\n';
      continue;
    }
    if (n <= (k - 1) * (k + 2) / 2) {
      cout << 0 << '\n';
      continue;
    }
    long long ans = 0;
    for (int i = k; i <= n; i++) {
      long long tmp = c[k][i];
      long long rem = n - i;
      long long x = (dp[n][rem][k] + dp[n][rem][k - 1]) % mod;
      tmp *= x;
      tmp %= mod;
      tmp *= fact[k];
      tmp %= mod;
      ans += tmp;
      ans %= mod;
    }
    cout << ans << '\n';
  }
}

### 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 T = 2 * 1e5 + 10, N = 1010, mod = 1e9 + 7, K = 50;
int dp[N][N][K], c[N][N];
long long fact[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  c[0][0] = 1;
  for (int i = 1; i < N; i++) {
    c[0][i] = 1;
    c[i][i] = 1;
    for (int j = 1; j < i; j++) {
      c[j][i] = c[j - 1][i - 1] + c[j][i - 1];
      c[j][i] %= mod;
    }
  }
  fact[0] = 1;
  for (int i = 1; i < N; i++) {
    fact[i] = fact[i - 1] * i;
    fact[i] %= mod;
  }
  dp[0][0][0] = 1;
  for (int i = 1; i < N; i++) dp[i][0][0] = 1;
  for (int i = 1; i < N; i++) {
    for (int j = 0; j < N; j++) {
      for (int k = 1; k < K; k++) {
        dp[i][j][k] = dp[i - 1][j][k];
        if (j >= i) dp[i][j][k] += dp[i - 1][j - i][k - 1];
        dp[i][j][k] %= mod;
      }
    }
  }
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    long long n, k;
    cin >> n >> k;
    if (k >= 50) {
      cout << 0 << '\n';
      continue;
    }
    if (n <= (k - 1) * (k + 2) / 2) {
      cout << 0 << '\n';
      continue;
    }
    long long ans = 0;
    for (int i = k; i <= n; i++) {
      long long tmp = c[k][i];
      long long rem = n - i;
      long long x = (dp[n][rem][k] + dp[n][rem][k - 1]) % mod;
      tmp *= x;
      tmp %= mod;
      tmp *= fact[k];
      tmp %= mod;
      ans += tmp;
      ans %= mod;
    }
    cout << ans << '\n';
  }
}

```