#include <bits/stdc++.h>
using namespace std;
long long s[1005][1005];
long long fact[1005];
long long invfact[1005];
long long choose[1005][1005];
long long s1(int x, int y) {
  if (y < 0) return 0;
  return s[x][y];
}
long long sum(int k, int n) {
  long long sum1 = 0;
  for (int i = 0; i <= n; ++i) {
    sum1 = (sum1 + choose[n - i][k] * s1(k, i));
    if (sum1 > 1E18) sum1 %= 1000000007;
  }
  return sum1;
}
int main() {
  fact[0] = 1;
  for (int i = 1; i <= 1000; ++i) fact[i] = (i * fact[i - 1]) % 1000000007;
  invfact[0] = 1;
  for (long long i = 1; i <= 1000; ++i) {
    long long l = 1;
    while (l % i != 0) l += 1000000007;
    l /= i;
    invfact[i] = (l * invfact[i - 1]) % 1000000007;
  }
  s[0][0] = 1;
  for (int i = 1; i <= 1000; ++i) s[0][i] = 0;
  for (int i = 1; i <= 1000; ++i) {
    for (int j = 0; j <= 1000; ++j) {
      s[i][j] = (i * s1(i - 1, j - i + 1) + s1(i, j - i)) % 1000000007;
    }
  }
  for (int i = 0; i <= 1000; ++i)
    for (int j = 0; j <= 1000; ++j) {
      if (i < j) {
        choose[i][j] = 0;
        continue;
      }
      choose[i][j] =
          ((fact[i] * invfact[i - j]) % 1000000007 * invfact[j]) % 1000000007;
    }
  int t;
  cin >> t;
  for (int i = 0; i < t; ++i) {
    int k, n;
    cin >> n >> k;
    printf("%d\n", sum(k, n) % 1000000007);
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 s[1005][1005];
long long fact[1005];
long long invfact[1005];
long long choose[1005][1005];
long long s1(int x, int y) {
  if (y < 0) return 0;
  return s[x][y];
}
long long sum(int k, int n) {
  long long sum1 = 0;
  for (int i = 0; i <= n; ++i) {
    sum1 = (sum1 + choose[n - i][k] * s1(k, i));
    if (sum1 > 1E18) sum1 %= 1000000007;
  }
  return sum1;
}
int main() {
  fact[0] = 1;
  for (int i = 1; i <= 1000; ++i) fact[i] = (i * fact[i - 1]) % 1000000007;
  invfact[0] = 1;
  for (long long i = 1; i <= 1000; ++i) {
    long long l = 1;
    while (l % i != 0) l += 1000000007;
    l /= i;
    invfact[i] = (l * invfact[i - 1]) % 1000000007;
  }
  s[0][0] = 1;
  for (int i = 1; i <= 1000; ++i) s[0][i] = 0;
  for (int i = 1; i <= 1000; ++i) {
    for (int j = 0; j <= 1000; ++j) {
      s[i][j] = (i * s1(i - 1, j - i + 1) + s1(i, j - i)) % 1000000007;
    }
  }
  for (int i = 0; i <= 1000; ++i)
    for (int j = 0; j <= 1000; ++j) {
      if (i < j) {
        choose[i][j] = 0;
        continue;
      }
      choose[i][j] =
          ((fact[i] * invfact[i - j]) % 1000000007 * invfact[j]) % 1000000007;
    }
  int t;
  cin >> t;
  for (int i = 0; i < t; ++i) {
    int k, n;
    cin >> n >> k;
    printf("%d\n", sum(k, n) % 1000000007);
  }
}

```