#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 10;
const int K = 50 + 10;
const int MOD = 1e9 + 7;
int t, n, k;
int dp[K][N][N];
int fac[N];
int chs[N][N];
int add(int a, int b) {
  long long x = (a + b) % MOD;
  return x;
}
int main() {
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      if (i == j || i == 0) {
        chs[i][j] = 1;
        continue;
      }
      if (i > j) {
        chs[i][j] = 0;
        continue;
      }
      if (i == 1) {
        chs[i][j] = j;
        continue;
      }
      chs[i][j] = add(chs[i - 1][j - 1], chs[i][j - 1]);
    }
  }
  fac[0] = 1;
  for (int i = 1; i < N; i++) {
    long long help = fac[i - 1];
    help = (help * i) % MOD;
    fac[i] = help;
  }
  for (int j = 1; j <= N; j++)
    for (int x = j; x <= N; x++) dp[1][j][x] = 1;
  for (int i = 2; i < K; i++)
    for (int j = 1; j < N; j++)
      for (int x = 1; x < N; x++) {
        if (x >= 2) dp[i][j][x] = add(dp[i][j][x], dp[i][j][x - 1]);
        if (j > x && x >= 2)
          dp[i][j][x] = add(dp[i][j][x], dp[i - 1][j - x][x - 1]);
      }
  scanf("%d", &t);
  for (int o = 0; o < t; o++) {
    scanf("%d %d", &n, &k);
    if (chs[2][k + 1] > n) {
      printf("%d\n", 0);
      continue;
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      long long h = dp[k][i][n];
      h = (h * chs[k][n - i + k]) % MOD;
      int h2 = h;
      ans = add(ans, h2);
    }
    ans = (ans * fac[k]) % MOD;
    cout << ans << endl;
  }
}

### 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;
const int N = 1e3 + 10;
const int K = 50 + 10;
const int MOD = 1e9 + 7;
int t, n, k;
int dp[K][N][N];
int fac[N];
int chs[N][N];
int add(int a, int b) {
  long long x = (a + b) % MOD;
  return x;
}
int main() {
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      if (i == j || i == 0) {
        chs[i][j] = 1;
        continue;
      }
      if (i > j) {
        chs[i][j] = 0;
        continue;
      }
      if (i == 1) {
        chs[i][j] = j;
        continue;
      }
      chs[i][j] = add(chs[i - 1][j - 1], chs[i][j - 1]);
    }
  }
  fac[0] = 1;
  for (int i = 1; i < N; i++) {
    long long help = fac[i - 1];
    help = (help * i) % MOD;
    fac[i] = help;
  }
  for (int j = 1; j <= N; j++)
    for (int x = j; x <= N; x++) dp[1][j][x] = 1;
  for (int i = 2; i < K; i++)
    for (int j = 1; j < N; j++)
      for (int x = 1; x < N; x++) {
        if (x >= 2) dp[i][j][x] = add(dp[i][j][x], dp[i][j][x - 1]);
        if (j > x && x >= 2)
          dp[i][j][x] = add(dp[i][j][x], dp[i - 1][j - x][x - 1]);
      }
  scanf("%d", &t);
  for (int o = 0; o < t; o++) {
    scanf("%d %d", &n, &k);
    if (chs[2][k + 1] > n) {
      printf("%d\n", 0);
      continue;
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      long long h = dp[k][i][n];
      h = (h * chs[k][n - i + k]) % MOD;
      int h2 = h;
      ans = add(ans, h2);
    }
    ans = (ans * fac[k]) % MOD;
    cout << ans << endl;
  }
}

```