#include <bits/stdc++.h>
using namespace std;
int pd[1024][1024][50], pd2[1024][1024];
int solve(int x, int falta, int ele) {
  if (falta == 0 && ele == 0) return 1;
  if (falta < 0 || ele == 0 || x > falta) return 0;
  int &ret = pd[x][falta][ele];
  if (ret != -1) return ret;
  ret = solve(x + 1, falta, ele) + solve(x + 1, falta - x, ele - 1);
  if (ret >= 1000000007) ret -= 1000000007;
  return ret;
}
int solve2(int espaco, int ele) {
  if (ele == 0) return 1;
  if (espaco == 0) return 0;
  int &ret = pd2[espaco][ele];
  if (ret != -1) return ret;
  ret = solve2(espaco, ele - 1) + solve2(espaco - 1, ele);
  if (ret >= 1000000007) ret -= 1000000007;
  return ret;
}
long long fat[1024];
int main() {
  int t;
  cin >> t;
  fat[0] = 1;
  for (int i = 1; i < 1024; i++) {
    fat[i] = (long long)i * (fat[i - 1]);
    fat[i] %= 1000000007;
  }
  memset(pd, -1, sizeof pd), memset(pd2, -1, sizeof pd2);
  while (t--) {
    int n, k;
    cin >> n >> k;
    long long ans = 0;
    for (int e = (-k + k * k) / 2; e + k <= n; e++) {
      ans += ((long long)solve(0, e, k) * (long long)solve2(k + 1, n - k - e)) %
             1000000007;
      if (ans >= 1000000007) ans -= 1000000007;
    }
    ans *= fat[k];
    ans %= 1000000007;
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Please formulate 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;
int pd[1024][1024][50], pd2[1024][1024];
int solve(int x, int falta, int ele) {
  if (falta == 0 && ele == 0) return 1;
  if (falta < 0 || ele == 0 || x > falta) return 0;
  int &ret = pd[x][falta][ele];
  if (ret != -1) return ret;
  ret = solve(x + 1, falta, ele) + solve(x + 1, falta - x, ele - 1);
  if (ret >= 1000000007) ret -= 1000000007;
  return ret;
}
int solve2(int espaco, int ele) {
  if (ele == 0) return 1;
  if (espaco == 0) return 0;
  int &ret = pd2[espaco][ele];
  if (ret != -1) return ret;
  ret = solve2(espaco, ele - 1) + solve2(espaco - 1, ele);
  if (ret >= 1000000007) ret -= 1000000007;
  return ret;
}
long long fat[1024];
int main() {
  int t;
  cin >> t;
  fat[0] = 1;
  for (int i = 1; i < 1024; i++) {
    fat[i] = (long long)i * (fat[i - 1]);
    fat[i] %= 1000000007;
  }
  memset(pd, -1, sizeof pd), memset(pd2, -1, sizeof pd2);
  while (t--) {
    int n, k;
    cin >> n >> k;
    long long ans = 0;
    for (int e = (-k + k * k) / 2; e + k <= n; e++) {
      ans += ((long long)solve(0, e, k) * (long long)solve2(k + 1, n - k - e)) %
             1000000007;
      if (ans >= 1000000007) ans -= 1000000007;
    }
    ans *= fat[k];
    ans %= 1000000007;
    cout << ans << endl;
  }
  return 0;
}

```