#include <bits/stdc++.h>
using namespace std;
int f[51][1010][1010], P = 1000000007, n, m, K, c[1060][51];
long long ans[51][1010], M[51];
void read(int &x) {
  char ch = getchar();
  int mark = 1;
  for (; ch != '-' && (ch < '0' || ch > '9'); ch = getchar())
    ;
  if (ch == '-') mark = -1, ch = getchar();
  for (x = 0; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - 48;
  x *= mark;
}
int main() {
  read(m);
  for (int k = 0; k <= 1000; k++) f[0][0][k] = 1;
  for (int i = 1; i <= 50; i++)
    for (int j = 0; j <= 1000; j++)
      for (int k = 1; k <= 1000; k++) {
        f[i][j][k] = f[i][j][k - 1];
        if (j >= k) (f[i][j][k] += f[i - 1][j - k][k - 1]) %= P;
      }
  M[1] = 1;
  for (int i = 2; i <= 50; i++) M[i] = M[i - 1] * i % P;
  c[0][0] = 1;
  for (int i = 1; i <= 1050; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= min(i, 50); j++)
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % P;
  }
  for (int i = 1; i <= 50; i++)
    for (int j = 1; j <= 1000; j++) {
      for (int k = 1; k <= j; k++)
        (ans[i][j] += (long long)f[i][k][1000] * c[j - k + i][i]) %= P;
      (ans[i][j] *= M[i]) %= P;
    }
  for (int i = 1; i <= m; i++) {
    read(n);
    read(K);
    if (n >= K * (K + 1) / 2)
      printf("%d\n", ans[K][n]);
    else
      printf("0\n");
  }
  return 0;
}

### Prompt

Your task is to create 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 f[51][1010][1010], P = 1000000007, n, m, K, c[1060][51];
long long ans[51][1010], M[51];
void read(int &x) {
  char ch = getchar();
  int mark = 1;
  for (; ch != '-' && (ch < '0' || ch > '9'); ch = getchar())
    ;
  if (ch == '-') mark = -1, ch = getchar();
  for (x = 0; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - 48;
  x *= mark;
}
int main() {
  read(m);
  for (int k = 0; k <= 1000; k++) f[0][0][k] = 1;
  for (int i = 1; i <= 50; i++)
    for (int j = 0; j <= 1000; j++)
      for (int k = 1; k <= 1000; k++) {
        f[i][j][k] = f[i][j][k - 1];
        if (j >= k) (f[i][j][k] += f[i - 1][j - k][k - 1]) %= P;
      }
  M[1] = 1;
  for (int i = 2; i <= 50; i++) M[i] = M[i - 1] * i % P;
  c[0][0] = 1;
  for (int i = 1; i <= 1050; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= min(i, 50); j++)
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % P;
  }
  for (int i = 1; i <= 50; i++)
    for (int j = 1; j <= 1000; j++) {
      for (int k = 1; k <= j; k++)
        (ans[i][j] += (long long)f[i][k][1000] * c[j - k + i][i]) %= P;
      (ans[i][j] *= M[i]) %= P;
    }
  for (int i = 1; i <= m; i++) {
    read(n);
    read(K);
    if (n >= K * (K + 1) / 2)
      printf("%d\n", ans[K][n]);
    else
      printf("0\n");
  }
  return 0;
}

```