#include <bits/stdc++.h>
using namespace std;
long long int fact[1006];
int dp[1002][1002][48];
int solve(int sum, int limit, int box) {
  if (box == 0) return sum >= 0;
  if (limit < 1 || sum < 1) return 0;
  if (dp[sum][limit][box] != -1) {
    return dp[sum][limit][box];
  }
  int ret = 0;
  ret = solve(sum - limit, limit - 1, box - 1);
  ret += solve(sum, limit - 1, box);
  if (ret >= ((long long int)1e9 + 7)) ret -= ((long long int)1e9 + 7);
  ret += solve(sum - 1, limit, box);
  if (ret >= ((long long int)1e9 + 7)) ret -= ((long long int)1e9 + 7);
  ret -= solve(sum - 1, limit - 1, box);
  if (ret < 0) ret += ((long long int)1e9 + 7);
  dp[sum][limit][box] = ret;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  fact[0] = 1;
  for (int i = 1; i < 1005; i++) {
    fact[i] = (fact[i - 1] * i) % ((long long int)1e9 + 7);
  }
  int t;
  scanf("%d", &t);
  memset(dp, -1, sizeof(dp));
  while (t--) {
    int n, k;
    long long int ans;
    scanf("%d%d", &n, &k);
    if (k >= 48) {
      printf("0\n");
      continue;
    }
    ans = solve(n, n, k);
    ans = (ans * fact[k]) % ((long long int)1e9 + 7);
    ans = (ans + ((long long int)1e9 + 7)) % ((long long int)1e9 + 7);
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Please formulate 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 int fact[1006];
int dp[1002][1002][48];
int solve(int sum, int limit, int box) {
  if (box == 0) return sum >= 0;
  if (limit < 1 || sum < 1) return 0;
  if (dp[sum][limit][box] != -1) {
    return dp[sum][limit][box];
  }
  int ret = 0;
  ret = solve(sum - limit, limit - 1, box - 1);
  ret += solve(sum, limit - 1, box);
  if (ret >= ((long long int)1e9 + 7)) ret -= ((long long int)1e9 + 7);
  ret += solve(sum - 1, limit, box);
  if (ret >= ((long long int)1e9 + 7)) ret -= ((long long int)1e9 + 7);
  ret -= solve(sum - 1, limit - 1, box);
  if (ret < 0) ret += ((long long int)1e9 + 7);
  dp[sum][limit][box] = ret;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  fact[0] = 1;
  for (int i = 1; i < 1005; i++) {
    fact[i] = (fact[i - 1] * i) % ((long long int)1e9 + 7);
  }
  int t;
  scanf("%d", &t);
  memset(dp, -1, sizeof(dp));
  while (t--) {
    int n, k;
    long long int ans;
    scanf("%d%d", &n, &k);
    if (k >= 48) {
      printf("0\n");
      continue;
    }
    ans = solve(n, n, k);
    ans = (ans * fact[k]) % ((long long int)1e9 + 7);
    ans = (ans + ((long long int)1e9 + 7)) % ((long long int)1e9 + 7);
    printf("%lld\n", ans);
  }
  return 0;
}

```