#include <bits/stdc++.h>
using namespace std;
long long inv[1001];
long long fact[1001];
long long invfact[1001];
long long sum[1001][1001];
int main() {
  int t, m = 1000000007, i, j;
  inv[1] = 1;
  for (i = 2; i <= 1000; i++) inv[i] = inv[m % i] * (m - m / i) % m;
  fact[0] = invfact[0] = 1;
  for (i = 1; i <= 1000; i++) fact[i] = fact[i - 1] * i % m;
  for (i = 1; i <= 1000; i++) invfact[i] = invfact[i - 1] * inv[i] % m;
  sum[1][0] = 1;
  for (i = 1; i <= 1000; i++) {
    for (j = 0; j <= 1000; j++) {
      if (sum[i][j] == 0) continue;
      if (i + j <= 1000) sum[i][i + j] = (sum[i][i + j] + sum[i][j]) % m;
      if (i < 1000) sum[i + 1][j] = (sum[i + 1][j] + sum[i][j]) % m;
    }
  }
  scanf("%d", &t);
  for (i = 0; i < t; i++) {
    int n, k, x;
    long long ans = 0;
    scanf("%d %d", &n, &k);
    x = n - (k * (k - 1) / 2 + (k - 1)) - 1;
    if (x < 0) {
      puts("0");
    } else {
      for (j = 0; j <= x; j++) {
        ans = (ans + sum[k][j] * fact[k + x - j] % m * invfact[x - j] % m) % m;
      }
      printf("%I64d\n", ans);
    }
  }
  return 0;
}

### 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 inv[1001];
long long fact[1001];
long long invfact[1001];
long long sum[1001][1001];
int main() {
  int t, m = 1000000007, i, j;
  inv[1] = 1;
  for (i = 2; i <= 1000; i++) inv[i] = inv[m % i] * (m - m / i) % m;
  fact[0] = invfact[0] = 1;
  for (i = 1; i <= 1000; i++) fact[i] = fact[i - 1] * i % m;
  for (i = 1; i <= 1000; i++) invfact[i] = invfact[i - 1] * inv[i] % m;
  sum[1][0] = 1;
  for (i = 1; i <= 1000; i++) {
    for (j = 0; j <= 1000; j++) {
      if (sum[i][j] == 0) continue;
      if (i + j <= 1000) sum[i][i + j] = (sum[i][i + j] + sum[i][j]) % m;
      if (i < 1000) sum[i + 1][j] = (sum[i + 1][j] + sum[i][j]) % m;
    }
  }
  scanf("%d", &t);
  for (i = 0; i < t; i++) {
    int n, k, x;
    long long ans = 0;
    scanf("%d %d", &n, &k);
    x = n - (k * (k - 1) / 2 + (k - 1)) - 1;
    if (x < 0) {
      puts("0");
    } else {
      for (j = 0; j <= x; j++) {
        ans = (ans + sum[k][j] * fact[k + x - j] % m * invfact[x - j] % m) % m;
      }
      printf("%I64d\n", ans);
    }
  }
  return 0;
}

```