#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f, N = 1010, P = 1e9 + 7;
template <class T>
inline bool chkmin(T &A, T B) {
  return B < A ? A = B, 1 : 0;
}
template <class T>
inline bool chkmax(T &A, T B) {
  return A < B ? A = B, 1 : 0;
}
template <class T, class _>
inline void Add(T &first, _ second) {
  first += second;
  if (first >= P) first -= P;
  if (first < 0) first += P;
}
int n, K;
long long C[N][N], Fac[N], dp[N][N];
int main() {
  int cas;
  Fac[0] = 1;
  for (int i = (1), i_end = (N - 10); i <= i_end; ++i)
    Fac[i] = Fac[i - 1] * i % P;
  for (int i = (0), i_end = (N - 10); i <= i_end; ++i) {
    C[i][0] = 1;
    for (int j = (1), j_end = (i); j <= j_end; ++j)
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % P;
  }
  dp[0][0] = 1;
  for (int i = (1), i_end = (N - 10); i <= i_end; ++i)
    for (int j = (1), j_end = (i); j <= j_end; ++j)
      dp[i][j] = (dp[i - j][j] + dp[i - j][j - 1]) % P;
  scanf("%d", &cas);
  while (cas--) {
    long long ans = 0;
    scanf("%d%d", &n, &K);
    for (int i = (K * (K + 1) / 2), i_end = (n); i <= i_end; ++i)
      Add(ans, dp[i][K] * C[n - i + K][K] % P);
    printf("%lld\n", ans * Fac[K] % P);
  }
  return 0;
}

### Prompt

Your task is to create 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 INF = 0x3f3f3f3f, N = 1010, P = 1e9 + 7;
template <class T>
inline bool chkmin(T &A, T B) {
  return B < A ? A = B, 1 : 0;
}
template <class T>
inline bool chkmax(T &A, T B) {
  return A < B ? A = B, 1 : 0;
}
template <class T, class _>
inline void Add(T &first, _ second) {
  first += second;
  if (first >= P) first -= P;
  if (first < 0) first += P;
}
int n, K;
long long C[N][N], Fac[N], dp[N][N];
int main() {
  int cas;
  Fac[0] = 1;
  for (int i = (1), i_end = (N - 10); i <= i_end; ++i)
    Fac[i] = Fac[i - 1] * i % P;
  for (int i = (0), i_end = (N - 10); i <= i_end; ++i) {
    C[i][0] = 1;
    for (int j = (1), j_end = (i); j <= j_end; ++j)
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % P;
  }
  dp[0][0] = 1;
  for (int i = (1), i_end = (N - 10); i <= i_end; ++i)
    for (int j = (1), j_end = (i); j <= j_end; ++j)
      dp[i][j] = (dp[i - j][j] + dp[i - j][j - 1]) % P;
  scanf("%d", &cas);
  while (cas--) {
    long long ans = 0;
    scanf("%d%d", &n, &K);
    for (int i = (K * (K + 1) / 2), i_end = (n); i <= i_end; ++i)
      Add(ans, dp[i][K] * C[n - i + K][K] % P);
    printf("%lld\n", ans * Fac[K] % P);
  }
  return 0;
}

```