#include <bits/stdc++.h>
using namespace std;
const int maxn = (int)(1e3 + 111);
const int inf = (int)(1e9 + 7);
const int base = (int)(1e9 + 7);
const int mod = (int)(1e9 + 7);
const double eps = (double)(1e-9);
const int R = (int)(1e3 + 10);
int t, n, k;
long long first[55], dp[2][maxn][maxn], sum[maxn][maxn], c[maxn][maxn];
long long add(long long a, long long b) {
  long long res = a + b;
  if (res >= mod) res -= mod;
  return res;
}
void cnk() {
  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 - 1] + c[i - 1][j]) % mod;
    }
  }
}
void precalc() {
  first[0] = 1;
  for (int i = 1; i <= 50; ++i) first[i] = (first[i - 1] * i) % mod;
  dp[0][0][1] = 1;
  for (int i = 0; i < 50; ++i) {
    for (int j = 0; j < R; ++j) {
      for (int mx = 0; mx < R; ++mx) {
        dp[0][j][mx + 1] = add(dp[0][j][mx + 1], dp[0][j][mx]);
        if (j + mx < R)
          dp[1][j + mx][mx + 1] = add(dp[0][j][mx], dp[1][j + mx][mx + 1]);
        if (mx == R - 1) sum[i][j] = (dp[0][j][mx] * first[i]) % mod;
      }
    }
    for (int j = 0; j < R; ++j) {
      for (int mx = 0; mx < R; dp[1][j][mx] = 0, ++mx)
        dp[0][j][mx] = dp[1][j][mx];
    }
  }
}
int main() {
  ios_base ::sync_with_stdio(false);
  precalc();
  cnk();
  cin >> t;
  for (int cs = 1; cs <= t; ++cs) {
    cin >> n >> k;
    if (k > 50) {
      cout << 0 << endl;
      continue;
    }
    long long ans = 0;
    for (int size = 0; size <= n; ++size) {
      if (n - size + k < 0) break;
      ans = add(ans, (sum[k][size] * c[n - size + k][k]) % 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 maxn = (int)(1e3 + 111);
const int inf = (int)(1e9 + 7);
const int base = (int)(1e9 + 7);
const int mod = (int)(1e9 + 7);
const double eps = (double)(1e-9);
const int R = (int)(1e3 + 10);
int t, n, k;
long long first[55], dp[2][maxn][maxn], sum[maxn][maxn], c[maxn][maxn];
long long add(long long a, long long b) {
  long long res = a + b;
  if (res >= mod) res -= mod;
  return res;
}
void cnk() {
  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 - 1] + c[i - 1][j]) % mod;
    }
  }
}
void precalc() {
  first[0] = 1;
  for (int i = 1; i <= 50; ++i) first[i] = (first[i - 1] * i) % mod;
  dp[0][0][1] = 1;
  for (int i = 0; i < 50; ++i) {
    for (int j = 0; j < R; ++j) {
      for (int mx = 0; mx < R; ++mx) {
        dp[0][j][mx + 1] = add(dp[0][j][mx + 1], dp[0][j][mx]);
        if (j + mx < R)
          dp[1][j + mx][mx + 1] = add(dp[0][j][mx], dp[1][j + mx][mx + 1]);
        if (mx == R - 1) sum[i][j] = (dp[0][j][mx] * first[i]) % mod;
      }
    }
    for (int j = 0; j < R; ++j) {
      for (int mx = 0; mx < R; dp[1][j][mx] = 0, ++mx)
        dp[0][j][mx] = dp[1][j][mx];
    }
  }
}
int main() {
  ios_base ::sync_with_stdio(false);
  precalc();
  cnk();
  cin >> t;
  for (int cs = 1; cs <= t; ++cs) {
    cin >> n >> k;
    if (k > 50) {
      cout << 0 << endl;
      continue;
    }
    long long ans = 0;
    for (int size = 0; size <= n; ++size) {
      if (n - size + k < 0) break;
      ans = add(ans, (sum[k][size] * c[n - size + k][k]) % mod);
    }
    cout << ans << endl;
  }
  return 0;
}

```