#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9 + 10, MAX = 2000, MOD = 1e9 + 7;
void OUT(long double o, int x) {
  cout << fixed << setprecision(x) << o;
  return;
}
long long dp3[MAX][MAX], dp4[MAX][MAX], dpx4[MAX][MAX], p[MAX];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  p[0] = 1;
  for (int i = 1; i <= 1000; i++) p[i] = (p[i - 1] * i) % MOD;
  for (int i = 1; i <= 1000; i++)
    dp3[1][i] = 1, dpx4[1][i] = i, dp4[1][i] = i, dpx4[0][i] = 1, dp4[0][i] = 1;
  dpx4[0][0] = 1, dp4[0][0] = 1;
  for (long long d = 1; d <= 1000; d++) {
    for (long long k = 1; k <= d + 1; k++) {
      for (int n = max(d + 1, (k * (k + 1)) / 2); n <= 1000; n++) {
        dp3[k][n] = (dp3[k][n] + dpx4[k - 1][n - d - 1]) % MOD;
        dp4[k][n] = (dp4[k][n - 1] + dp3[k][n]) % MOD;
      }
    }
    for (long long k = 1; k <= d + 1; k++) {
      for (int n = max(d + 1, (k * (k + 1)) / 2); n <= 1000; n++)
        dpx4[k][n] = dp4[k][n];
    }
  }
  long long t, n, k;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    cout << (dp4[k][n] * p[k]) % MOD << "\n";
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 INF = 1e9 + 10, MAX = 2000, MOD = 1e9 + 7;
void OUT(long double o, int x) {
  cout << fixed << setprecision(x) << o;
  return;
}
long long dp3[MAX][MAX], dp4[MAX][MAX], dpx4[MAX][MAX], p[MAX];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  p[0] = 1;
  for (int i = 1; i <= 1000; i++) p[i] = (p[i - 1] * i) % MOD;
  for (int i = 1; i <= 1000; i++)
    dp3[1][i] = 1, dpx4[1][i] = i, dp4[1][i] = i, dpx4[0][i] = 1, dp4[0][i] = 1;
  dpx4[0][0] = 1, dp4[0][0] = 1;
  for (long long d = 1; d <= 1000; d++) {
    for (long long k = 1; k <= d + 1; k++) {
      for (int n = max(d + 1, (k * (k + 1)) / 2); n <= 1000; n++) {
        dp3[k][n] = (dp3[k][n] + dpx4[k - 1][n - d - 1]) % MOD;
        dp4[k][n] = (dp4[k][n - 1] + dp3[k][n]) % MOD;
      }
    }
    for (long long k = 1; k <= d + 1; k++) {
      for (int n = max(d + 1, (k * (k + 1)) / 2); n <= 1000; n++)
        dpx4[k][n] = dp4[k][n];
    }
  }
  long long t, n, k;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    cout << (dp4[k][n] * p[k]) % MOD << "\n";
  }
  return 0;
}

```