#include <bits/stdc++.h>
using namespace std;
void setIO(string St) {
  string IN = St + ".in", OUT = St + ".out";
  freopen(IN.c_str(), "r", stdin);
  freopen(OUT.c_str(), "w", stdout);
}
const int MOD = 1000000007;
int Z[1008];
int jc[2008], C[2008][2008], Ans[1008][50], F[1008][50][1008];
int main() {
  jc[1] = 1;
  for (int i = 2; i <= 2000; i++)
    jc[i] = (long long)jc[i - 1] * (long long)i % (long long)MOD;
  int t = 1, s = 1;
  for (int i = 1; i <= 1000; i++) {
    if (i > s) t++, s += t;
    Z[i] = t;
  }
  C[1][0] = C[1][1] = 1;
  for (int i = 2; i <= 2000; i++) {
    C[i][0] = 1;
    for (int j = 1; j <= i; j++)
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % MOD;
  }
  F[0][0][0] = 1;
  for (int i = 1; i <= 1000; i++) {
    for (int k = 1; k <= Z[i]; k++)
      for (int L = 1; L <= i; L++) {
        F[i][k][L] +=
            (F[i - L][k - 1][min(L - 1, i - L)] + F[i][k][L - 1]) % MOD;
        F[i][k][L] %= MOD;
      }
  }
  for (int n = 1; n <= 1000; n++)
    for (int k = 1; k <= Z[n]; k++) {
      int s = 0;
      for (int i = 1; i <= n; i++) {
        int t =
            (long long)F[i][k][i] * (long long)C[n - i + k][k] % (long long)MOD;
        s = (s + t) % MOD;
      }
      Ans[n][k] = (long long)s * (long long)jc[k] % (long long)MOD;
    }
  int nq;
  scanf("%d", &nq);
  for (; nq; nq--) {
    int n, k;
    scanf("%d%d", &n, &k);
    if (k > 45)
      printf("0\n");
    else
      printf("%d\n", Ans[n][k]);
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
void setIO(string St) {
  string IN = St + ".in", OUT = St + ".out";
  freopen(IN.c_str(), "r", stdin);
  freopen(OUT.c_str(), "w", stdout);
}
const int MOD = 1000000007;
int Z[1008];
int jc[2008], C[2008][2008], Ans[1008][50], F[1008][50][1008];
int main() {
  jc[1] = 1;
  for (int i = 2; i <= 2000; i++)
    jc[i] = (long long)jc[i - 1] * (long long)i % (long long)MOD;
  int t = 1, s = 1;
  for (int i = 1; i <= 1000; i++) {
    if (i > s) t++, s += t;
    Z[i] = t;
  }
  C[1][0] = C[1][1] = 1;
  for (int i = 2; i <= 2000; i++) {
    C[i][0] = 1;
    for (int j = 1; j <= i; j++)
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % MOD;
  }
  F[0][0][0] = 1;
  for (int i = 1; i <= 1000; i++) {
    for (int k = 1; k <= Z[i]; k++)
      for (int L = 1; L <= i; L++) {
        F[i][k][L] +=
            (F[i - L][k - 1][min(L - 1, i - L)] + F[i][k][L - 1]) % MOD;
        F[i][k][L] %= MOD;
      }
  }
  for (int n = 1; n <= 1000; n++)
    for (int k = 1; k <= Z[n]; k++) {
      int s = 0;
      for (int i = 1; i <= n; i++) {
        int t =
            (long long)F[i][k][i] * (long long)C[n - i + k][k] % (long long)MOD;
        s = (s + t) % MOD;
      }
      Ans[n][k] = (long long)s * (long long)jc[k] % (long long)MOD;
    }
  int nq;
  scanf("%d", &nq);
  for (; nq; nq--) {
    int n, k;
    scanf("%d%d", &n, &k);
    if (k > 45)
      printf("0\n");
    else
      printf("%d\n", Ans[n][k]);
  }
  return 0;
}

```