#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
const int mod = 1000000007;
int dp[N][50][N];
int c[N][N];
long long fac[N];
int f[N][N];
void add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
void init() {
  for (int i = 0; i < N; i++) {
    c[i][0] = c[i][i] = 1;
    for (int j = 1; j < i; j++) {
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
    }
  }
  dp[0][1][0] = dp[0][0][0] = 1;
  for (int i = 0; i < N - 1; i++) {
    for (int j = 0; j < 50; j++) {
      for (int k = 0; k < N; k++) {
        if (!dp[i][j][k]) continue;
        add(dp[i + 1][j][k], dp[i][j][k]);
        if (k + i + 1 < N && j + 1 < 50)
          add(dp[i + 1][j + 1][k + i + 1], dp[i][j][k]);
      }
    }
  }
  fac[0] = 1;
  for (int i = 1; i < N; i++) fac[i] = fac[i - 1] * i % mod;
  for (int i = 1; i < N; i++) f[1][i] = 1;
  for (int i = 2; i < N; i++) {
    f[i][i] = f[i - 1][i - 1];
    int sum = (f[i - 1][i - 1] + f[i - 1][i]) % mod;
    for (int j = i + 1; j < N; j++) {
      f[i][j] = sum;
      add(sum, f[i - 1][j]);
    }
  }
  for (int i = 1; i < N; i++)
    for (int j = 1; j < N; j++) add(f[i][j], f[i][j - 1]);
}
void solve() {
  int n, m;
  cin >> n >> m;
  if (m * (m - 1) / 2 > n) {
    cout << 0 << endl;
    return;
  }
  long long ans = 0;
  for (int i = 0; i <= n; i++) {
    if (!dp[n - 1][m][i]) continue;
    ans = (ans + (long long)f[m][n - i] * dp[n - 1][m][i]) % mod;
  }
  ans = ans * fac[m] % mod;
  cout << ans << endl;
}
int main() {
  init();
  int t;
  ios::sync_with_stdio(false);
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Please formulate 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 N = 1010;
const int mod = 1000000007;
int dp[N][50][N];
int c[N][N];
long long fac[N];
int f[N][N];
void add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
void init() {
  for (int i = 0; i < N; i++) {
    c[i][0] = c[i][i] = 1;
    for (int j = 1; j < i; j++) {
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
    }
  }
  dp[0][1][0] = dp[0][0][0] = 1;
  for (int i = 0; i < N - 1; i++) {
    for (int j = 0; j < 50; j++) {
      for (int k = 0; k < N; k++) {
        if (!dp[i][j][k]) continue;
        add(dp[i + 1][j][k], dp[i][j][k]);
        if (k + i + 1 < N && j + 1 < 50)
          add(dp[i + 1][j + 1][k + i + 1], dp[i][j][k]);
      }
    }
  }
  fac[0] = 1;
  for (int i = 1; i < N; i++) fac[i] = fac[i - 1] * i % mod;
  for (int i = 1; i < N; i++) f[1][i] = 1;
  for (int i = 2; i < N; i++) {
    f[i][i] = f[i - 1][i - 1];
    int sum = (f[i - 1][i - 1] + f[i - 1][i]) % mod;
    for (int j = i + 1; j < N; j++) {
      f[i][j] = sum;
      add(sum, f[i - 1][j]);
    }
  }
  for (int i = 1; i < N; i++)
    for (int j = 1; j < N; j++) add(f[i][j], f[i][j - 1]);
}
void solve() {
  int n, m;
  cin >> n >> m;
  if (m * (m - 1) / 2 > n) {
    cout << 0 << endl;
    return;
  }
  long long ans = 0;
  for (int i = 0; i <= n; i++) {
    if (!dp[n - 1][m][i]) continue;
    ans = (ans + (long long)f[m][n - i] * dp[n - 1][m][i]) % mod;
  }
  ans = ans * fac[m] % mod;
  cout << ans << endl;
}
int main() {
  init();
  int t;
  ios::sync_with_stdio(false);
  cin >> t;
  while (t--) solve();
  return 0;
}

```