#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const int N = 1005;
long long f[N][N], F[N][N], binom[N][N], fac[N];
void prepro() {
  f[0][0] = 1;
  for (int n = (1); n <= (1001); ++n)
    for (int k = (1); k <= (50); ++k) {
      for (int i = k; i <= n; i += k)
        f[n][k] = (f[n - i][k - 1] + f[n][k]) % mod;
    }
  binom[0][0] = 1;
  for (int n = (1); n <= (1001); ++n)
    for (int k = (0); k <= (n); ++k) {
      if (k == 0 || k == n)
        binom[n][k] = 1;
      else
        binom[n][k] = (binom[n - 1][k] + binom[n - 1][k - 1]) % mod;
    }
  fac[0] = 1;
  for (int n = (1); n <= (1001); ++n) fac[n] = (fac[n - 1] * n) % mod;
  F[0][0] = 1;
  for (int n = (0); n <= (1000); ++n)
    for (int k = (1); k <= (50); ++k)
      F[n][k] = ((f[n][k] + f[n][k - 1]) * fac[k]) % mod;
}
long long BINOM(int a, int b) {
  if (a < b) return 0;
  if (b < 0) return 0;
  return binom[a][b];
}
long long jebaj() {
  int n, k;
  cin >> n >> k;
  if (k > 45) return 0;
  long long wyn = 0;
  for (int a = (0); a <= ((n + 1) - 1); ++a) {
    wyn = (wyn + F[a][k] * BINOM(n - a, k)) % mod;
  }
  return wyn;
}
int main() {
  prepro();
  ios_base::sync_with_stdio(0);
  int t;
  cin >> t;
  for (int i = (0); i <= ((t)-1); ++i) {
    cout << jebaj() << "\n";
  }
}

### 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;
const long long mod = 1000000007;
const int N = 1005;
long long f[N][N], F[N][N], binom[N][N], fac[N];
void prepro() {
  f[0][0] = 1;
  for (int n = (1); n <= (1001); ++n)
    for (int k = (1); k <= (50); ++k) {
      for (int i = k; i <= n; i += k)
        f[n][k] = (f[n - i][k - 1] + f[n][k]) % mod;
    }
  binom[0][0] = 1;
  for (int n = (1); n <= (1001); ++n)
    for (int k = (0); k <= (n); ++k) {
      if (k == 0 || k == n)
        binom[n][k] = 1;
      else
        binom[n][k] = (binom[n - 1][k] + binom[n - 1][k - 1]) % mod;
    }
  fac[0] = 1;
  for (int n = (1); n <= (1001); ++n) fac[n] = (fac[n - 1] * n) % mod;
  F[0][0] = 1;
  for (int n = (0); n <= (1000); ++n)
    for (int k = (1); k <= (50); ++k)
      F[n][k] = ((f[n][k] + f[n][k - 1]) * fac[k]) % mod;
}
long long BINOM(int a, int b) {
  if (a < b) return 0;
  if (b < 0) return 0;
  return binom[a][b];
}
long long jebaj() {
  int n, k;
  cin >> n >> k;
  if (k > 45) return 0;
  long long wyn = 0;
  for (int a = (0); a <= ((n + 1) - 1); ++a) {
    wyn = (wyn + F[a][k] * BINOM(n - a, k)) % mod;
  }
  return wyn;
}
int main() {
  prepro();
  ios_base::sync_with_stdio(0);
  int t;
  cin >> t;
  for (int i = (0); i <= ((t)-1); ++i) {
    cout << jebaj() << "\n";
  }
}

```