#include <bits/stdc++.h>
using namespace std;
int dp[1010][1010][47];
long long fact[1010];
long long moduloexp(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (1LL * res * a) % 1000000007;
    b /= 2;
    a = (1LL * a * a) % 1000000007;
  }
  return res;
}
void init() {
  fact[0] = 1;
  for (int i = 1; i < 1010; i++) fact[i] = (fact[i - 1] * i) % 1000000007;
}
int fun(int a, int b, int c) {
  if (c > 32) return 0;
  if (b < 0) return 0;
  if (c == 0) {
    if (b == 0) return 1;
    return 0;
  }
  if (a > b) return 0;
  if (dp[a][b][c] != -1) return dp[a][b][c];
  return dp[a][b][c] =
             (fun(a + 1, b - a, c - 1) + fun(a + 1, b, c)) % 1000000007;
}
int comb[1010][1010];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  init();
  for (int i = 0; i < 1010; i++) dp[i][0][0] = 1;
  for (int i = 0; i < 1010; i++) comb[i][0] = 1;
  for (int i = 1; i < 1010; i++) {
    for (int j = 0; j < 1010; j++)
      if (j)
        comb[i][j] = (comb[i - 1][j - 1] + comb[i - 1][j]) % 1000000007;
      else
        comb[i][j] = comb[i - 1][j];
  }
  for (int i = 1008; i >= 0; i--) {
    for (int j = 1008; j >= 0; j--) {
      for (int k = 46; k >= 0; k--) {
        if (i > j) continue;
        dp[i][j][k] = (dp[i + 1][j][k] + dp[i + 1][j - i][k - 1]) % 1000000007;
        ;
      }
    }
  }
  while (t--) {
    int k, n;
    cin >> n >> k;
    long long ans = 0;
    for (int i = 0; i <= n - k; i++) {
      if (k < 47)
        ans = (ans + (dp[0][i][k] * fact[k] % 1000000007 * comb[n - i][k] %
                      1000000007)) %
              1000000007;
    }
    cout << ans << '\n';
  }
}

### 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 dp[1010][1010][47];
long long fact[1010];
long long moduloexp(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (1LL * res * a) % 1000000007;
    b /= 2;
    a = (1LL * a * a) % 1000000007;
  }
  return res;
}
void init() {
  fact[0] = 1;
  for (int i = 1; i < 1010; i++) fact[i] = (fact[i - 1] * i) % 1000000007;
}
int fun(int a, int b, int c) {
  if (c > 32) return 0;
  if (b < 0) return 0;
  if (c == 0) {
    if (b == 0) return 1;
    return 0;
  }
  if (a > b) return 0;
  if (dp[a][b][c] != -1) return dp[a][b][c];
  return dp[a][b][c] =
             (fun(a + 1, b - a, c - 1) + fun(a + 1, b, c)) % 1000000007;
}
int comb[1010][1010];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  init();
  for (int i = 0; i < 1010; i++) dp[i][0][0] = 1;
  for (int i = 0; i < 1010; i++) comb[i][0] = 1;
  for (int i = 1; i < 1010; i++) {
    for (int j = 0; j < 1010; j++)
      if (j)
        comb[i][j] = (comb[i - 1][j - 1] + comb[i - 1][j]) % 1000000007;
      else
        comb[i][j] = comb[i - 1][j];
  }
  for (int i = 1008; i >= 0; i--) {
    for (int j = 1008; j >= 0; j--) {
      for (int k = 46; k >= 0; k--) {
        if (i > j) continue;
        dp[i][j][k] = (dp[i + 1][j][k] + dp[i + 1][j - i][k - 1]) % 1000000007;
        ;
      }
    }
  }
  while (t--) {
    int k, n;
    cin >> n >> k;
    long long ans = 0;
    for (int i = 0; i <= n - k; i++) {
      if (k < 47)
        ans = (ans + (dp[0][i][k] * fact[k] % 1000000007 * comb[n - i][k] %
                      1000000007)) %
              1000000007;
    }
    cout << ans << '\n';
  }
}

```