#include <bits/stdc++.h>
using namespace std;
template <class T>
inline bool chkmin(T& a, T const& b) {
  return a > b ? a = b, true : false;
}
template <class T>
inline bool chkmax(T& a, T const& b) {
  return a < b ? a = b, true : false;
}
const int M = 1005, K = 300, N = M + K;
int C[N][N], dp[M][K], ans[M][M];
int _, n, k;
int main() {
  for (int i = (0), i_end__ = (N); i < i_end__; ++i)
    for (int j = (C[i][0] = 1), j_end__ = (i + 1); j < j_end__; ++j)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % 1000000007;
  dp[0][0] = 1;
  for (int i = (1), i_end__ = (M); i < i_end__; ++i)
    for (int j = (M)-1, j_end__ = (0); j >= j_end__; --j)
      for (int k = (K)-1, k_end__ = (0); k >= k_end__; --k)
        if (i + j < M && dp[j][k]) {
          dp[j + i][k + 1] =
              (dp[j + i][k + 1] + 1LL * (k + 1) * dp[j][k]) % 1000000007;
        }
  for (int i = (1), i_end__ = (M); i < i_end__; ++i)
    for (int k = (1), k_end__ = (K); k < k_end__; ++k)
      for (int j = (1), j_end__ = (i + 1); j < j_end__; ++j)
        if (dp[j][k]) {
          ans[i][k] =
              (ans[i][k] + 1LL * dp[j][k] * C[i - j + k][k]) % 1000000007;
        }
  for (scanf("%d", &_); _; --_) {
    scanf("%d %d", &n, &k);
    printf("%d\n", ans[n][k]);
  }
  return 0;
}

### 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;
template <class T>
inline bool chkmin(T& a, T const& b) {
  return a > b ? a = b, true : false;
}
template <class T>
inline bool chkmax(T& a, T const& b) {
  return a < b ? a = b, true : false;
}
const int M = 1005, K = 300, N = M + K;
int C[N][N], dp[M][K], ans[M][M];
int _, n, k;
int main() {
  for (int i = (0), i_end__ = (N); i < i_end__; ++i)
    for (int j = (C[i][0] = 1), j_end__ = (i + 1); j < j_end__; ++j)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % 1000000007;
  dp[0][0] = 1;
  for (int i = (1), i_end__ = (M); i < i_end__; ++i)
    for (int j = (M)-1, j_end__ = (0); j >= j_end__; --j)
      for (int k = (K)-1, k_end__ = (0); k >= k_end__; --k)
        if (i + j < M && dp[j][k]) {
          dp[j + i][k + 1] =
              (dp[j + i][k + 1] + 1LL * (k + 1) * dp[j][k]) % 1000000007;
        }
  for (int i = (1), i_end__ = (M); i < i_end__; ++i)
    for (int k = (1), k_end__ = (K); k < k_end__; ++k)
      for (int j = (1), j_end__ = (i + 1); j < j_end__; ++j)
        if (dp[j][k]) {
          ans[i][k] =
              (ans[i][k] + 1LL * dp[j][k] * C[i - j + k][k]) % 1000000007;
        }
  for (scanf("%d", &_); _; --_) {
    scanf("%d %d", &n, &k);
    printf("%d\n", ans[n][k]);
  }
  return 0;
}

```