#include <bits/stdc++.h>
using namespace std;
const long long md = 1e9 + 7;
int t;
int dp[52][1005][1005];
int dp2[52][1005];
int c[1501][60];
long long f[100];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  dp[0][0][1] = 1;
  f[0] = 1;
  for (long long i = 1; i <= 50; ++i) f[i] = (f[i - 1] * i) % md;
  for (int i = 0; i <= 50; ++i) {
    for (int j = 0; j <= 1001; ++j) {
      long long sum = 0;
      for (int k = 0; k <= 1001; ++k) {
        sum = sum + dp[i][j][k];
        if (sum > md) sum -= md;
        if (j + k <= 1000) {
          dp[i + 1][j + k][k + 1] = dp[i + 1][j + k][k + 1] + sum;
          if (dp[i + 1][j + k][k + 1] > md) dp[i + 1][j + k][k + 1] -= md;
        }
      }
      for (int k = 1; k <= 1001; ++k) {
        dp2[i][j] = (dp2[i][j] + dp[i][j][k]);
        if (dp2[i][j] > md) dp2[i][j] -= md;
      }
      dp2[i][j] = (dp2[i][j] * f[i]) % md;
    }
  }
  c[0][0] = 1;
  for (int i = 1; i <= 1500; ++i)
    for (int j = 0; j <= min(55, i); ++j) {
      c[i][j] = c[i - 1][j];
      if (j) c[i][j] += c[i - 1][j - 1];
      if (c[i][j] >= md) c[i][j] -= md;
    }
  for (int i = 1; i <= t; ++i) {
    int x, y;
    cin >> x >> y;
    long long res = 0;
    if (y > 50) {
      cout << "0" << '\n';
      continue;
    }
    for (int i = 0; i <= x; ++i)
      res = (res +
             (long long)((long long)dp2[y][i] * (long long)c[x - i + y][y]) %
                 md) %
            md;
    cout << res << '\n';
  }
}

### Prompt

Generate 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 long long md = 1e9 + 7;
int t;
int dp[52][1005][1005];
int dp2[52][1005];
int c[1501][60];
long long f[100];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  dp[0][0][1] = 1;
  f[0] = 1;
  for (long long i = 1; i <= 50; ++i) f[i] = (f[i - 1] * i) % md;
  for (int i = 0; i <= 50; ++i) {
    for (int j = 0; j <= 1001; ++j) {
      long long sum = 0;
      for (int k = 0; k <= 1001; ++k) {
        sum = sum + dp[i][j][k];
        if (sum > md) sum -= md;
        if (j + k <= 1000) {
          dp[i + 1][j + k][k + 1] = dp[i + 1][j + k][k + 1] + sum;
          if (dp[i + 1][j + k][k + 1] > md) dp[i + 1][j + k][k + 1] -= md;
        }
      }
      for (int k = 1; k <= 1001; ++k) {
        dp2[i][j] = (dp2[i][j] + dp[i][j][k]);
        if (dp2[i][j] > md) dp2[i][j] -= md;
      }
      dp2[i][j] = (dp2[i][j] * f[i]) % md;
    }
  }
  c[0][0] = 1;
  for (int i = 1; i <= 1500; ++i)
    for (int j = 0; j <= min(55, i); ++j) {
      c[i][j] = c[i - 1][j];
      if (j) c[i][j] += c[i - 1][j - 1];
      if (c[i][j] >= md) c[i][j] -= md;
    }
  for (int i = 1; i <= t; ++i) {
    int x, y;
    cin >> x >> y;
    long long res = 0;
    if (y > 50) {
      cout << "0" << '\n';
      continue;
    }
    for (int i = 0; i <= x; ++i)
      res = (res +
             (long long)((long long)dp2[y][i] * (long long)c[x - i + y][y]) %
                 md) %
            md;
    cout << res << '\n';
  }
}

```