#include <bits/stdc++.h>
using namespace std;
const int maxN = 1000 + 5;
const long long mod = 1000 * 1000 * 1000 + 7;
long long c[maxN][maxN];
long long fact[maxN];
long long add(long long x, long long y) { return (x + y) % mod; }
long long mul(long long x, long long y) { return (x * y) % mod; }
long long b[maxN][maxN];
int main() {
  ios_base::sync_with_stdio(false);
  c[0][0] = 1;
  for (int i = 1; i < maxN; i++) {
    c[i][i] = c[i][0] = 1;
    for (int j = 1; j < i; j++) c[i][j] = add(c[i - 1][j - 1], c[i - 1][j]);
  }
  int N = maxN - 2;
  fact[0] = 1;
  for (int i = 1; i < maxN; i++) fact[i] = mul(fact[i - 1], i);
  b[0][1] = 1;
  b[1][2] = 1;
  for (int i = 1; i <= N; i++)
    for (int k = 1; k <= i; k++)
      b[i][k] = add(b[i - k][k], b[i - k + 1][k - 1]);
  for (int i = 1; i <= N; i++)
    for (int k = 1; k <= N; k++) b[i][k] = mul(b[i][k], fact[k]);
  int T;
  cin >> T;
  while (T--) {
    int n, k;
    cin >> n >> k;
    if (n - k < 0 || k * (k - 1) / 2 > n - k) {
      cout << 0 << endl;
      continue;
    }
    long long ans = 0;
    for (int i = 0; i <= n; i++) ans = (ans + b[i][k] * c[n - i][k]) % mod;
    cout << ans << "\n";
  }
  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 maxN = 1000 + 5;
const long long mod = 1000 * 1000 * 1000 + 7;
long long c[maxN][maxN];
long long fact[maxN];
long long add(long long x, long long y) { return (x + y) % mod; }
long long mul(long long x, long long y) { return (x * y) % mod; }
long long b[maxN][maxN];
int main() {
  ios_base::sync_with_stdio(false);
  c[0][0] = 1;
  for (int i = 1; i < maxN; i++) {
    c[i][i] = c[i][0] = 1;
    for (int j = 1; j < i; j++) c[i][j] = add(c[i - 1][j - 1], c[i - 1][j]);
  }
  int N = maxN - 2;
  fact[0] = 1;
  for (int i = 1; i < maxN; i++) fact[i] = mul(fact[i - 1], i);
  b[0][1] = 1;
  b[1][2] = 1;
  for (int i = 1; i <= N; i++)
    for (int k = 1; k <= i; k++)
      b[i][k] = add(b[i - k][k], b[i - k + 1][k - 1]);
  for (int i = 1; i <= N; i++)
    for (int k = 1; k <= N; k++) b[i][k] = mul(b[i][k], fact[k]);
  int T;
  cin >> T;
  while (T--) {
    int n, k;
    cin >> n >> k;
    if (n - k < 0 || k * (k - 1) / 2 > n - k) {
      cout << 0 << endl;
      continue;
    }
    long long ans = 0;
    for (int i = 0; i <= n; i++) ans = (ans + b[i][k] * c[n - i][k]) % mod;
    cout << ans << "\n";
  }
  return 0;
}

```