#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const int mxN = 1010;
long long dp[mxN][mxN];
long long mem[mxN][mxN];
long long fact[mxN];
int main(int argc, char** argv) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  dp[0][0] = 1;
  for (int j = 1; j < mxN; j++) {
    for (int k = j; k >= 1; k--) {
      for (int n = k * (k - 1) / 2 + j; n < mxN; n++) {
        dp[k][n] += dp[k - 1][n - j];
        if (dp[k][n] >= mod) dp[k][n] -= mod;
      }
    }
  }
  fact[0] = fact[1] = 1;
  for (int i = 2; i < mxN; i++) fact[i] = fact[i - 1] * i % mod;
  mem[0][0] = 1;
  for (int i = 1; i < mxN; i++) {
    mem[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      mem[i][j] = mem[i - 1][j] + mem[i - 1][j - 1];
      if (mem[i][j] >= mod) mem[i][j] -= mod;
    }
  }
  for (int k = 1; k < mxN; k++) {
    for (int n = mxN - 1; n >= k * (k + 1) / 2; n--) {
      for (int j = k * (k + 1) / 2; j < n; j++) {
        dp[k][n] = (dp[k][n] + dp[k][j] * mem[n - j + k][k]) % mod;
      }
    }
  }
  int T;
  cin >> T;
  while (T--) {
    int N, K;
    cin >> N >> K;
    if (N < K * (K - 1) / 2) {
      cout << "0\n";
      continue;
    }
    cout << dp[K][N] * fact[K] % mod << '\n';
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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 long long mod = 1000000007;
const int mxN = 1010;
long long dp[mxN][mxN];
long long mem[mxN][mxN];
long long fact[mxN];
int main(int argc, char** argv) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  dp[0][0] = 1;
  for (int j = 1; j < mxN; j++) {
    for (int k = j; k >= 1; k--) {
      for (int n = k * (k - 1) / 2 + j; n < mxN; n++) {
        dp[k][n] += dp[k - 1][n - j];
        if (dp[k][n] >= mod) dp[k][n] -= mod;
      }
    }
  }
  fact[0] = fact[1] = 1;
  for (int i = 2; i < mxN; i++) fact[i] = fact[i - 1] * i % mod;
  mem[0][0] = 1;
  for (int i = 1; i < mxN; i++) {
    mem[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      mem[i][j] = mem[i - 1][j] + mem[i - 1][j - 1];
      if (mem[i][j] >= mod) mem[i][j] -= mod;
    }
  }
  for (int k = 1; k < mxN; k++) {
    for (int n = mxN - 1; n >= k * (k + 1) / 2; n--) {
      for (int j = k * (k + 1) / 2; j < n; j++) {
        dp[k][n] = (dp[k][n] + dp[k][j] * mem[n - j + k][k]) % mod;
      }
    }
  }
  int T;
  cin >> T;
  while (T--) {
    int N, K;
    cin >> N >> K;
    if (N < K * (K - 1) / 2) {
      cout << "0\n";
      continue;
    }
    cout << dp[K][N] * fact[K] % mod << '\n';
  }
  return 0;
}

```