#include <bits/stdc++.h>
using namespace std;
long long dp[1105][55], binomni_koeficijent[1005][55], resenje[1005][55],
    faktorijal[1005];
long long BK(long long x, long long y) {
  if (x < 0) return 0;
  return binomni_koeficijent[x][y];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t, n, k, i, j, S, br, ans = 0, z = 0, mod = 1000000007;
  cin >> t;
  dp[0][0] = 1;
  for (S = 1; S <= 1000; S++) {
    for (i = 1; i <= 50; i++) {
      br = 1;
      while (S - i * br >= 0) {
        dp[S][i] = (dp[S][i] + dp[S - i * br][i - 1]) % mod;
        br++;
      }
    }
  }
  for (i = 0; i <= 1000; i++) binomni_koeficijent[i][0] = 1;
  for (i = 1; i <= 1000; i++)
    for (j = 1; j <= 50; j++)
      binomni_koeficijent[i][j] =
          (binomni_koeficijent[i - 1][j - 1] + binomni_koeficijent[i - 1][j]) %
          mod;
  faktorijal[0] = 1;
  for (i = 1; i <= 50; i++) faktorijal[i] = (faktorijal[i - 1] * i) % mod;
  for (n = 1; n <= 1000; n++)
    for (k = 1; k <= 50; k++) {
      for (i = n; i >= 0; i--) {
        z = z + BK(n - i - 1, k - 1);
        z = z % mod;
        j = (faktorijal[k] * dp[i + k][k]) % mod;
        ans = ans + (j * z) % mod;
      }
      resenje[n][k] = ans % mod;
      ans = 0;
      z = 0;
    }
  while (t--) {
    cin >> n >> k;
    if (k > 50)
      cout << "0" << endl;
    else
      cout << resenje[n][k] << endl;
  }
  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;
long long dp[1105][55], binomni_koeficijent[1005][55], resenje[1005][55],
    faktorijal[1005];
long long BK(long long x, long long y) {
  if (x < 0) return 0;
  return binomni_koeficijent[x][y];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t, n, k, i, j, S, br, ans = 0, z = 0, mod = 1000000007;
  cin >> t;
  dp[0][0] = 1;
  for (S = 1; S <= 1000; S++) {
    for (i = 1; i <= 50; i++) {
      br = 1;
      while (S - i * br >= 0) {
        dp[S][i] = (dp[S][i] + dp[S - i * br][i - 1]) % mod;
        br++;
      }
    }
  }
  for (i = 0; i <= 1000; i++) binomni_koeficijent[i][0] = 1;
  for (i = 1; i <= 1000; i++)
    for (j = 1; j <= 50; j++)
      binomni_koeficijent[i][j] =
          (binomni_koeficijent[i - 1][j - 1] + binomni_koeficijent[i - 1][j]) %
          mod;
  faktorijal[0] = 1;
  for (i = 1; i <= 50; i++) faktorijal[i] = (faktorijal[i - 1] * i) % mod;
  for (n = 1; n <= 1000; n++)
    for (k = 1; k <= 50; k++) {
      for (i = n; i >= 0; i--) {
        z = z + BK(n - i - 1, k - 1);
        z = z % mod;
        j = (faktorijal[k] * dp[i + k][k]) % mod;
        ans = ans + (j * z) % mod;
      }
      resenje[n][k] = ans % mod;
      ans = 0;
      z = 0;
    }
  while (t--) {
    cin >> n >> k;
    if (k > 50)
      cout << "0" << endl;
    else
      cout << resenje[n][k] << endl;
  }
  return 0;
}

```