#include <bits/stdc++.h>
using namespace std;
const int INF = (1 << 30) - 1;
const long long lINF = (1LL << 62) - 1;
const double dINF = 1e40;
const int MAXN = 1010, MAXP = 50, mod = 1000000007;
int f[MAXN][MAXN][MAXP];
int C[MAXN][MAXN], fac[MAXN];
void getMod(int &v) { v >= mod ? v -= mod : 0; }
int main() {
  const int r = 1000;
  f[0][0][0] = f[0][0][1] = 1;
  for (int m = 1; m <= r; m++) {
    f[m][0][0] = f[m][0][1] = 1;
    for (int i = 1; i <= r; i++)
      for (int j = 1; j * (j - 1) / 2 <= i; j++) {
        f[m][i][j] = f[m - 1][i][j] + (i >= m ? f[m - 1][i - m][j - 1] : 0);
        getMod(f[m][i][j]);
      }
  }
  C[0][0] = 1;
  for (int i = 1; i <= r; i++) {
    C[i][0] = 1;
    for (int j = 1; j <= i; j++)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mod;
  }
  fac[0] = 1;
  for (int i = 1; i <= r; i++) fac[i] = (long long)fac[i - 1] * i % mod;
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, m;
    scanf("%d%d", &n, &m);
    int ans = 0;
    for (int sv = m * (m - 1) / 2; sv + m <= n; sv++) {
      ans = (ans + (long long)((long long)C[n - sv][m] * fac[m] % mod) *
                       f[r][sv][m]) %
            mod;
    }
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 = (1 << 30) - 1;
const long long lINF = (1LL << 62) - 1;
const double dINF = 1e40;
const int MAXN = 1010, MAXP = 50, mod = 1000000007;
int f[MAXN][MAXN][MAXP];
int C[MAXN][MAXN], fac[MAXN];
void getMod(int &v) { v >= mod ? v -= mod : 0; }
int main() {
  const int r = 1000;
  f[0][0][0] = f[0][0][1] = 1;
  for (int m = 1; m <= r; m++) {
    f[m][0][0] = f[m][0][1] = 1;
    for (int i = 1; i <= r; i++)
      for (int j = 1; j * (j - 1) / 2 <= i; j++) {
        f[m][i][j] = f[m - 1][i][j] + (i >= m ? f[m - 1][i - m][j - 1] : 0);
        getMod(f[m][i][j]);
      }
  }
  C[0][0] = 1;
  for (int i = 1; i <= r; i++) {
    C[i][0] = 1;
    for (int j = 1; j <= i; j++)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mod;
  }
  fac[0] = 1;
  for (int i = 1; i <= r; i++) fac[i] = (long long)fac[i - 1] * i % mod;
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, m;
    scanf("%d%d", &n, &m);
    int ans = 0;
    for (int sv = m * (m - 1) / 2; sv + m <= n; sv++) {
      ans = (ans + (long long)((long long)C[n - sv][m] * fac[m] % mod) *
                       f[r][sv][m]) %
            mod;
    }
    printf("%d\n", ans);
  }
  return 0;
}

```