#include <bits/stdc++.h>
using namespace std;
const int mod = 1000 * 1000 * 1000 + 7;
const int inf = 1000 * 1000 * 1000 * 2;
const int rad = 400;
const int maxn = 1000 + 10;
int f[maxn];
int dp[maxn][maxn], c[maxn * 2][maxn * 2];
inline int solve(int n, int k) {
  if (k < 0) return 0;
  if (n == 0)
    if (k == 0)
      return 1;
    else
      return 0;
  int &ret = dp[n][k];
  if (ret != -1) return ret;
  ret = 0;
  return ret = (solve(n - 1, k - (n - 1)) + solve(n, k - n)) % mod;
}
inline int C(int k, int n) {
  if (k > n || k < 0) return 0;
  if (k == 0 || k == n) return 1;
  int &ret = c[k][n];
  if (ret != -1) return ret;
  ret = 0;
  return ret = (C(k - 1, n - 1) + C(k, n - 1)) % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  memset(dp, -1, sizeof dp);
  memset(c, -1, sizeof c);
  f[0] = 1;
  for (int i = 1; i < maxn; ++i) f[i] = (1LL * f[i - 1] * i) % mod;
  int q;
  cin >> q;
  while (q-- > 0) {
    int n, k;
    cin >> n >> k;
    long long ans = 0;
    for (int i = (k * (k - 1)) / 2; i <= n - k; ++i)
      ans += (((1LL * f[k] * solve(k, i)) % mod) * C(k, n - i)) % mod,
          ans %= mod;
    cout << ans << endl;
  }
  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;
const int mod = 1000 * 1000 * 1000 + 7;
const int inf = 1000 * 1000 * 1000 * 2;
const int rad = 400;
const int maxn = 1000 + 10;
int f[maxn];
int dp[maxn][maxn], c[maxn * 2][maxn * 2];
inline int solve(int n, int k) {
  if (k < 0) return 0;
  if (n == 0)
    if (k == 0)
      return 1;
    else
      return 0;
  int &ret = dp[n][k];
  if (ret != -1) return ret;
  ret = 0;
  return ret = (solve(n - 1, k - (n - 1)) + solve(n, k - n)) % mod;
}
inline int C(int k, int n) {
  if (k > n || k < 0) return 0;
  if (k == 0 || k == n) return 1;
  int &ret = c[k][n];
  if (ret != -1) return ret;
  ret = 0;
  return ret = (C(k - 1, n - 1) + C(k, n - 1)) % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  memset(dp, -1, sizeof dp);
  memset(c, -1, sizeof c);
  f[0] = 1;
  for (int i = 1; i < maxn; ++i) f[i] = (1LL * f[i - 1] * i) % mod;
  int q;
  cin >> q;
  while (q-- > 0) {
    int n, k;
    cin >> n >> k;
    long long ans = 0;
    for (int i = (k * (k - 1)) / 2; i <= n - k; ++i)
      ans += (((1LL * f[k] * solve(k, i)) % mod) * C(k, n - i)) % mod,
          ans %= mod;
    cout << ans << endl;
  }
  return 0;
}

```