#include <bits/stdc++.h>
int dr[] = {2, 2, -2, -2, 1, -1, 1, -1};
int dc[] = {1, -1, 1, -1, 2, 2, -2, -2};
int dr1[] = {0, 0, 0, 1, 1, 1, -1, -1, -1};
int dc1[] = {-1, 0, 1, -1, 0, 1, -1, 0, 1};
int dr2[] = {-1, 0, 0, 1};
int dc2[] = {-1, -1, -2, -1};
using namespace std;
int dp[1005][1005][46];
long long int fr[100005];
long long int sum(long long int x) {}
long long int cal(long long int x, long long int y, long long int k) {
  if (k == 0) return 1;
  if (x == 0 || y >= x) return 0;
  int &r = dp[x][y][k];
  if (r != -1) return r;
  r = 0;
  r = cal(x, y + 1, k);
  r += cal(x - 1, y, k);
  r %= 1000000007;
  if (y < x) r += cal(x - y - 1, y + 1, k - 1);
  r %= 1000000007;
  r = (r - cal(x - 1, y + 1, k) + 1000000007) % 1000000007;
  return r;
}
void pre() {
  long long int i, j;
  fr[0] = 1;
  for (i = 1; i < 1001; i++) {
    fr[i] = (fr[i - 1] * i) % 1000000007;
  }
}
int main() {
  long long int tc, n, k;
  memset(dp, -1, sizeof(dp));
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  pre();
  cin >> tc;
  while (tc--) {
    cin >> n >> k;
    if (k > 45) {
      cout << "0" << endl;
      continue;
    }
    long long int ans = cal(n, 0, k);
    ans = ans * fr[k];
    ans %= 1000000007;
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Develop a solution in Cpp 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>
int dr[] = {2, 2, -2, -2, 1, -1, 1, -1};
int dc[] = {1, -1, 1, -1, 2, 2, -2, -2};
int dr1[] = {0, 0, 0, 1, 1, 1, -1, -1, -1};
int dc1[] = {-1, 0, 1, -1, 0, 1, -1, 0, 1};
int dr2[] = {-1, 0, 0, 1};
int dc2[] = {-1, -1, -2, -1};
using namespace std;
int dp[1005][1005][46];
long long int fr[100005];
long long int sum(long long int x) {}
long long int cal(long long int x, long long int y, long long int k) {
  if (k == 0) return 1;
  if (x == 0 || y >= x) return 0;
  int &r = dp[x][y][k];
  if (r != -1) return r;
  r = 0;
  r = cal(x, y + 1, k);
  r += cal(x - 1, y, k);
  r %= 1000000007;
  if (y < x) r += cal(x - y - 1, y + 1, k - 1);
  r %= 1000000007;
  r = (r - cal(x - 1, y + 1, k) + 1000000007) % 1000000007;
  return r;
}
void pre() {
  long long int i, j;
  fr[0] = 1;
  for (i = 1; i < 1001; i++) {
    fr[i] = (fr[i - 1] * i) % 1000000007;
  }
}
int main() {
  long long int tc, n, k;
  memset(dp, -1, sizeof(dp));
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  pre();
  cin >> tc;
  while (tc--) {
    cin >> n >> k;
    if (k > 45) {
      cout << "0" << endl;
      continue;
    }
    long long int ans = cal(n, 0, k);
    ans = ans * fr[k];
    ans %= 1000000007;
    cout << ans << endl;
  }
  return 0;
}

```