#include <bits/stdc++.h>
using namespace std;
const int N = 2005;
const long long mod = 1000000007;
long long C[N][N];
long long dp[N][N];
long long factor[N];
int main() {
  ios::sync_with_stdio(false);
  int t, n, k;
  for (int i = 0; i < N; i++) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; j++) {
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mod;
    }
  }
  for (int i = 0; i < N; i++) dp[i][1] = 1;
  for (int i = 2; i < N; i++)
    for (int j = 0; j < N; j++) {
      if (j - i >= 0) dp[j][i] = dp[j - i][i];
      if (j - i + 1 >= 0) dp[j][i] += dp[j - i + 1][i - 1];
      if (dp[j][i] >= mod) dp[j][i] -= mod;
    }
  factor[0] = 1;
  for (int i = 1; i < N; i++) {
    factor[i] = factor[i - 1] * i;
    factor[i] %= mod;
  }
  cin >> t;
  while (t--) {
    cin >> n >> k;
    if (n < k) {
      puts("0");
      continue;
    }
    if (k * (k - 1) > 2 * n) {
      puts("0");
      continue;
    }
    long long ans = 0;
    for (int x = 0; x <= n; x++) {
      long long s = C[n - x][k] * dp[x][k] % mod;
      ans += s;
      if (ans >= mod) ans -= mod;
    }
    cout << ans * factor[k] % mod << endl;
  }
}

### Prompt

Please provide a CPP coded solution 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 N = 2005;
const long long mod = 1000000007;
long long C[N][N];
long long dp[N][N];
long long factor[N];
int main() {
  ios::sync_with_stdio(false);
  int t, n, k;
  for (int i = 0; i < N; i++) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; j++) {
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mod;
    }
  }
  for (int i = 0; i < N; i++) dp[i][1] = 1;
  for (int i = 2; i < N; i++)
    for (int j = 0; j < N; j++) {
      if (j - i >= 0) dp[j][i] = dp[j - i][i];
      if (j - i + 1 >= 0) dp[j][i] += dp[j - i + 1][i - 1];
      if (dp[j][i] >= mod) dp[j][i] -= mod;
    }
  factor[0] = 1;
  for (int i = 1; i < N; i++) {
    factor[i] = factor[i - 1] * i;
    factor[i] %= mod;
  }
  cin >> t;
  while (t--) {
    cin >> n >> k;
    if (n < k) {
      puts("0");
      continue;
    }
    if (k * (k - 1) > 2 * n) {
      puts("0");
      continue;
    }
    long long ans = 0;
    for (int x = 0; x <= n; x++) {
      long long s = C[n - x][k] * dp[x][k] % mod;
      ans += s;
      if (ans >= mod) ans -= mod;
    }
    cout << ans * factor[k] % mod << endl;
  }
}

```