#include <bits/stdc++.h>
using namespace std;
const int modul = 1000000007;
const int nmax = 1010;
const double e = 1e-8;
const double pi = acos(-1);
int n, m, stest, k;
int dp[nmax][nmax];
int res[nmax][nmax], c[nmax][nmax], gt[nmax];
void init() {
  dp[0][0] = 1;
  for (int i = 0; i <= 1000; i++)
    for (int j = 0; j <= i + 1; j++)
      if (i + j > 0) {
        if (i >= j - 1 && j >= 1)
          dp[i][j] = (dp[i][j] + dp[i - (j - 1)][j - 1]) % modul;
        if (i >= j) dp[i][j] = (dp[i][j] + dp[i - j][j]) % modul;
      }
  for (int i = 0; i <= 1000; i++)
    for (int j = 0; j <= i; j++)
      if (j == 0 || j == i)
        c[i][j] = 1;
      else
        c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % modul;
  gt[0] = 1;
  for (int i = 1; i <= 1000; i++) gt[i] = ((long long)gt[i - 1] * i) % modul;
  for (int i = 1; i <= 1000; i++) {
    int t = 1000;
    while (t >= 1 && dp[i - 1][t] == 0) t--;
    for (int j = 1; j <= t; j++) {
      res[i][j] = 0;
      for (int u = 0; u <= i - 1; u++) {
        int g = ((long long)dp[u][j] * c[i - u][j]) % modul;
        res[i][j] = (res[i][j] + g) % modul;
      }
      res[i][j] = ((long long)res[i][j] * gt[j]) % modul;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  init();
  for (cin >> stest; stest; stest--) {
    cin >> n >> k;
    cout << res[n][k] << endl;
  }
  return 0;
}

### 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 modul = 1000000007;
const int nmax = 1010;
const double e = 1e-8;
const double pi = acos(-1);
int n, m, stest, k;
int dp[nmax][nmax];
int res[nmax][nmax], c[nmax][nmax], gt[nmax];
void init() {
  dp[0][0] = 1;
  for (int i = 0; i <= 1000; i++)
    for (int j = 0; j <= i + 1; j++)
      if (i + j > 0) {
        if (i >= j - 1 && j >= 1)
          dp[i][j] = (dp[i][j] + dp[i - (j - 1)][j - 1]) % modul;
        if (i >= j) dp[i][j] = (dp[i][j] + dp[i - j][j]) % modul;
      }
  for (int i = 0; i <= 1000; i++)
    for (int j = 0; j <= i; j++)
      if (j == 0 || j == i)
        c[i][j] = 1;
      else
        c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % modul;
  gt[0] = 1;
  for (int i = 1; i <= 1000; i++) gt[i] = ((long long)gt[i - 1] * i) % modul;
  for (int i = 1; i <= 1000; i++) {
    int t = 1000;
    while (t >= 1 && dp[i - 1][t] == 0) t--;
    for (int j = 1; j <= t; j++) {
      res[i][j] = 0;
      for (int u = 0; u <= i - 1; u++) {
        int g = ((long long)dp[u][j] * c[i - u][j]) % modul;
        res[i][j] = (res[i][j] + g) % modul;
      }
      res[i][j] = ((long long)res[i][j] * gt[j]) % modul;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  init();
  for (cin >> stest; stest; stest--) {
    cin >> n >> k;
    cout << res[n][k] << endl;
  }
  return 0;
}

```