#include <bits/stdc++.h>
using namespace std;
const int maxn = 1111;
const int maxk = 57;
int n, k, t;
int f[maxk][maxn], c[maxn][maxn];
long long dp[maxk][maxn], frac[maxn];
void init() {
  memset(f, 0, sizeof(f));
  f[0][0] = 1;
  for (int i = 1, _c = 50; i <= _c; i++)
    for (int j = 1, _c = 1000; j <= _c; j++)
      if (j >= i) {
        f[i][j] = f[i][j - i] + f[i - 1][j - i];
        if (f[i][j] > 1000000007) f[i][j] -= 1000000007;
      } else
        f[i][j] = 0;
  memset(c, 0, sizeof(c));
  c[0][0] = 1;
  for (int i = 1, _c = 1050; i <= _c; i++) {
    c[i][0] = 1;
    for (int j = 1, _c = i; j <= _c; j++) {
      c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
      if (c[i][j] > 1000000007) c[i][j] -= 1000000007;
    }
  }
  frac[0] = 1;
  for (int i = 1, _c = 1000; i <= _c; i++)
    frac[i] = (frac[i - 1] * i) % 1000000007;
  for (int i = 1, _c = 50; i <= _c; i++)
    for (int j = 1, _c = 1000; j <= _c; j++) {
      dp[i][j] = 0;
      for (int k = 1, _c = j; k <= _c; k++)
        dp[i][j] =
            (dp[i][j] + 1ll * f[i][k] * c[j - k + i][j - k]) % 1000000007;
      dp[i][j] = (dp[i][j] * frac[i]) % 1000000007;
    }
}
int main() {
  init();
  scanf("%d", &t);
  for (int tt = 0, _a = (t); tt < _a; ++tt) {
    scanf("%d%d", &n, &k);
    if (k > 50)
      puts("0");
    else {
      printf("%I64d\n", dp[k][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;
const int maxn = 1111;
const int maxk = 57;
int n, k, t;
int f[maxk][maxn], c[maxn][maxn];
long long dp[maxk][maxn], frac[maxn];
void init() {
  memset(f, 0, sizeof(f));
  f[0][0] = 1;
  for (int i = 1, _c = 50; i <= _c; i++)
    for (int j = 1, _c = 1000; j <= _c; j++)
      if (j >= i) {
        f[i][j] = f[i][j - i] + f[i - 1][j - i];
        if (f[i][j] > 1000000007) f[i][j] -= 1000000007;
      } else
        f[i][j] = 0;
  memset(c, 0, sizeof(c));
  c[0][0] = 1;
  for (int i = 1, _c = 1050; i <= _c; i++) {
    c[i][0] = 1;
    for (int j = 1, _c = i; j <= _c; j++) {
      c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
      if (c[i][j] > 1000000007) c[i][j] -= 1000000007;
    }
  }
  frac[0] = 1;
  for (int i = 1, _c = 1000; i <= _c; i++)
    frac[i] = (frac[i - 1] * i) % 1000000007;
  for (int i = 1, _c = 50; i <= _c; i++)
    for (int j = 1, _c = 1000; j <= _c; j++) {
      dp[i][j] = 0;
      for (int k = 1, _c = j; k <= _c; k++)
        dp[i][j] =
            (dp[i][j] + 1ll * f[i][k] * c[j - k + i][j - k]) % 1000000007;
      dp[i][j] = (dp[i][j] * frac[i]) % 1000000007;
    }
}
int main() {
  init();
  scanf("%d", &t);
  for (int tt = 0, _a = (t); tt < _a; ++tt) {
    scanf("%d%d", &n, &k);
    if (k > 50)
      puts("0");
    else {
      printf("%I64d\n", dp[k][n]);
    }
  }
  return 0;
}

```