#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const int INF = 2e9;
const long long INF64 = 1e18;
const long double EPS = 1e-12;
const long double PI = 3.1415926535897932384626433832795028841;
const long long MD = 925262732;
const long long T = 634521;
const int N = 200100;
const int M = 301;
long long c[3000][3000], d[65][1001], ans[62][1001], p[1001];
int main() {
  p[0] = 1;
  for (long long i = 1; i <= 60; i++) p[i] = (p[i - 1] * i) % MOD;
  for (int i = 0; i <= 2000; i++)
    for (int j = 0; j <= i; j++)
      if (j == 0 || j == i)
        c[i][j] = 1;
      else
        c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % MOD;
  d[0][0] = 1;
  for (int z = 0; z <= 1000; z++)
    for (int i = 60; i >= 0; i--)
      for (int j = 0; j <= 1000; j++)
        if (j + z <= 1000) d[i + 1][j + z] = (d[i + 1][j + z] + d[i][j]) % MOD;
  for (int i = 1; i <= 60; i++)
    for (int j = 0; j <= 1000; j++)
      for (int u = 0; u <= j; u++)
        if (j - u >= i - 1)
          ans[i][j] = (ans[i][j] + (d[i][u] * c[j - u][i - 1])) % MOD;
  for (int i = 1; i <= 60; i++)
    for (int j = 1; j <= 1000; j++)
      ans[i][j] = (ans[i][j] + ans[i][j - 1]) % MOD;
  int t;
  cin >> t;
  for (int i = 0; i < int(t); i++) {
    int n, k;
    cin >> n >> k;
    if (k > 60) {
      printf("0\n");
      continue;
    }
    n--;
    long long sum = (ans[k][n] * p[k]) % MOD;
    printf("%I64d\n", sum);
  }
  return 0;
}

### 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 MOD = 1e9 + 7;
const int INF = 2e9;
const long long INF64 = 1e18;
const long double EPS = 1e-12;
const long double PI = 3.1415926535897932384626433832795028841;
const long long MD = 925262732;
const long long T = 634521;
const int N = 200100;
const int M = 301;
long long c[3000][3000], d[65][1001], ans[62][1001], p[1001];
int main() {
  p[0] = 1;
  for (long long i = 1; i <= 60; i++) p[i] = (p[i - 1] * i) % MOD;
  for (int i = 0; i <= 2000; i++)
    for (int j = 0; j <= i; j++)
      if (j == 0 || j == i)
        c[i][j] = 1;
      else
        c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % MOD;
  d[0][0] = 1;
  for (int z = 0; z <= 1000; z++)
    for (int i = 60; i >= 0; i--)
      for (int j = 0; j <= 1000; j++)
        if (j + z <= 1000) d[i + 1][j + z] = (d[i + 1][j + z] + d[i][j]) % MOD;
  for (int i = 1; i <= 60; i++)
    for (int j = 0; j <= 1000; j++)
      for (int u = 0; u <= j; u++)
        if (j - u >= i - 1)
          ans[i][j] = (ans[i][j] + (d[i][u] * c[j - u][i - 1])) % MOD;
  for (int i = 1; i <= 60; i++)
    for (int j = 1; j <= 1000; j++)
      ans[i][j] = (ans[i][j] + ans[i][j - 1]) % MOD;
  int t;
  cin >> t;
  for (int i = 0; i < int(t); i++) {
    int n, k;
    cin >> n >> k;
    if (k > 60) {
      printf("0\n");
      continue;
    }
    n--;
    long long sum = (ans[k][n] * p[k]) % MOD;
    printf("%I64d\n", sum);
  }
  return 0;
}

```