#include <bits/stdc++.h>
using namespace std;
template <class T>
inline bool chkmin(T &x, T y) {
  return y < x ? x = y, 1 : 0;
}
template <class T>
inline bool chkmax(T &x, T y) {
  return x < y ? x = y, 1 : 0;
}
inline long long Max(long long x, long long y) { return x > y ? x : y; }
inline long long Min(long long x, long long y) { return x < y ? x : y; }
int n, m;
long long fac[1002], inv[1002];
long long dp[47][1002], ans[1002][47];
void Add(long long &x, long long y) {
  x += y;
  if (x >= 1000000007) x -= 1000000007;
}
long long C(int n, int m) {
  if (n < 0 || m > n) return 0;
  return fac[n] * inv[m] % 1000000007 * inv[n - m] % 1000000007;
}
void Init() {
  fac[0] = fac[1] = 1;
  for (int i = (2), i_end_ = (1002); i < i_end_; i++)
    fac[i] = fac[i - 1] * i % 1000000007;
  inv[0] = inv[1] = 1;
  for (int i = (2), i_end_ = (1002); i < i_end_; i++)
    inv[i] = (1000000007 - (1000000007 / i)) * inv[1000000007 % i] % 1000000007;
  for (int i = (2), i_end_ = (1002); i < i_end_; i++)
    inv[i] = inv[i] * inv[i - 1] % 1000000007;
  dp[0][0] = 1;
  for (int i = (1), i_end_ = (47); i < i_end_; i++)
    for (int j = (i * (i + 1) / 2), j_end_ = (1002); j < j_end_; j++)
      Add(dp[i][j] = dp[i][j - i], dp[i - 1][j - i]);
  for (int i = (1), i_end_ = (1002); i < i_end_; i++) {
    for (int k = (1), k_end_ = (47); k < k_end_; k++) {
      for (int j = (k * (k + 1) / 2), j_end_ = (i); j <= j_end_; j++)
        Add(ans[i][k], C(i - j + k, k) * dp[k][j] % 1000000007);
      ans[i][k] = ans[i][k] * fac[k] % 1000000007;
    }
  }
}
int main() {
  Init();
  int cas;
  cin >> cas;
  while (cas--) {
    scanf("%d%d", &n, &m);
    printf("%I64d\n", m >= 45 ? 0 : ans[n][m]);
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
template <class T>
inline bool chkmin(T &x, T y) {
  return y < x ? x = y, 1 : 0;
}
template <class T>
inline bool chkmax(T &x, T y) {
  return x < y ? x = y, 1 : 0;
}
inline long long Max(long long x, long long y) { return x > y ? x : y; }
inline long long Min(long long x, long long y) { return x < y ? x : y; }
int n, m;
long long fac[1002], inv[1002];
long long dp[47][1002], ans[1002][47];
void Add(long long &x, long long y) {
  x += y;
  if (x >= 1000000007) x -= 1000000007;
}
long long C(int n, int m) {
  if (n < 0 || m > n) return 0;
  return fac[n] * inv[m] % 1000000007 * inv[n - m] % 1000000007;
}
void Init() {
  fac[0] = fac[1] = 1;
  for (int i = (2), i_end_ = (1002); i < i_end_; i++)
    fac[i] = fac[i - 1] * i % 1000000007;
  inv[0] = inv[1] = 1;
  for (int i = (2), i_end_ = (1002); i < i_end_; i++)
    inv[i] = (1000000007 - (1000000007 / i)) * inv[1000000007 % i] % 1000000007;
  for (int i = (2), i_end_ = (1002); i < i_end_; i++)
    inv[i] = inv[i] * inv[i - 1] % 1000000007;
  dp[0][0] = 1;
  for (int i = (1), i_end_ = (47); i < i_end_; i++)
    for (int j = (i * (i + 1) / 2), j_end_ = (1002); j < j_end_; j++)
      Add(dp[i][j] = dp[i][j - i], dp[i - 1][j - i]);
  for (int i = (1), i_end_ = (1002); i < i_end_; i++) {
    for (int k = (1), k_end_ = (47); k < k_end_; k++) {
      for (int j = (k * (k + 1) / 2), j_end_ = (i); j <= j_end_; j++)
        Add(ans[i][k], C(i - j + k, k) * dp[k][j] % 1000000007);
      ans[i][k] = ans[i][k] * fac[k] % 1000000007;
    }
  }
}
int main() {
  Init();
  int cas;
  cin >> cas;
  while (cas--) {
    scanf("%d%d", &n, &m);
    printf("%I64d\n", m >= 45 ? 0 : ans[n][m]);
  }
  return 0;
}

```