#include <bits/stdc++.h>
using namespace std;
long long ways[77][2100], ans[77][2100], c[2100][2100];
long long tests, a, b, fact[21000];
int main() {
  ios_base::sync_with_stdio(0);
  c[0][0] = 1;
  fact[0] = 1;
  for (int i = 1; i <= 1200; i++) fact[i] = fact[i - 1] * i % 1000000007;
  for (int i = 1; i <= 1000; i++)
    for (int j = 0; j <= i; j++)
      if (j == 0 || j == i)
        c[i][j] = 1;
      else
        c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % 1000000007;
  ways[0][0] = 1;
  for (int ad = 1; ad <= 1000; ad++) {
    for (int ttl = 60; ttl >= 0; ttl--)
      for (int old = 0; old <= 1000; old++) {
        ways[ttl + 1][old + ad] += ways[ttl][old];
        while (ways[ttl + 1][old + ad] >= 1000000007)
          ways[ttl + 1][old + ad] -= 1000000007;
      }
  }
  for (int ttl = 1; ttl <= 60; ttl++)
    for (int i = 0; i <= 1000; i++)
      for (int j = i; j <= 1000; j++)
        ans[ttl][j] =
            (ans[ttl][j] + ways[ttl][i] * c[j - i + ttl][ttl]) % 1000000007;
  for (int ttl = 1; ttl <= 60; ttl++)
    for (int i = 0; i <= 1000; i++)
      ans[ttl][i] = ans[ttl][i] * fact[ttl] % 1000000007;
  cin >> tests;
  for (; tests; --tests) {
    cin >> a >> b;
    swap(a, b);
    if (a > 65)
      cout << 0 << endl;
    else
      cout << ans[a][b] << endl;
  }
  cin.get();
  cin.get();
  return 0;
}

### Prompt

Your challenge is to write 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;
long long ways[77][2100], ans[77][2100], c[2100][2100];
long long tests, a, b, fact[21000];
int main() {
  ios_base::sync_with_stdio(0);
  c[0][0] = 1;
  fact[0] = 1;
  for (int i = 1; i <= 1200; i++) fact[i] = fact[i - 1] * i % 1000000007;
  for (int i = 1; i <= 1000; i++)
    for (int j = 0; j <= i; j++)
      if (j == 0 || j == i)
        c[i][j] = 1;
      else
        c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % 1000000007;
  ways[0][0] = 1;
  for (int ad = 1; ad <= 1000; ad++) {
    for (int ttl = 60; ttl >= 0; ttl--)
      for (int old = 0; old <= 1000; old++) {
        ways[ttl + 1][old + ad] += ways[ttl][old];
        while (ways[ttl + 1][old + ad] >= 1000000007)
          ways[ttl + 1][old + ad] -= 1000000007;
      }
  }
  for (int ttl = 1; ttl <= 60; ttl++)
    for (int i = 0; i <= 1000; i++)
      for (int j = i; j <= 1000; j++)
        ans[ttl][j] =
            (ans[ttl][j] + ways[ttl][i] * c[j - i + ttl][ttl]) % 1000000007;
  for (int ttl = 1; ttl <= 60; ttl++)
    for (int i = 0; i <= 1000; i++)
      ans[ttl][i] = ans[ttl][i] * fact[ttl] % 1000000007;
  cin >> tests;
  for (; tests; --tests) {
    cin >> a >> b;
    swap(a, b);
    if (a > 65)
      cout << 0 << endl;
    else
      cout << ans[a][b] << endl;
  }
  cin.get();
  cin.get();
  return 0;
}

```