#include <bits/stdc++.h>
using namespace std;
int dp[3][50][1002], fact[1050][50], ans[((int)(2e5 + 1))];
vector<pair<int, pair<int, int> > > v;
bool cmp(pair<int, pair<int, int> > a, pair<int, pair<int, int> > b) {
  if (a.first == b.first) return a.second.first > b.second.first;
  return a.first < b.first;
}
int main() {
  for (int i = 1; i <= 1046; i++) {
    fact[i][0] = 1;
    for (int j = 1; j <= min(i, 46); j++) {
      fact[i][j] = ((long long)fact[i][j - 1] * (i - j + 1)) % (1000000007);
    }
  }
  int t, n, k, tc;
  scanf("%d", &t);
  for (int i = 0; i < t; i++) {
    scanf("%d %d", &n, &k);
    v.push_back(make_pair(k, make_pair(n, i)));
  }
  sort(v.begin(), v.end(), cmp);
  bool cur;
  for (int i = 0; i < v.size(); i++) {
    k = v[i].first;
    n = v[i].second.first;
    if (k * (k + 1) > 2 * n) {
      ans[v[i].second.second] = 0;
      continue;
    }
    if (i == 0 || v[i].first != v[i - 1].first) {
      cur = 0;
      for (int rem = 0; rem <= 1000; rem++) {
        dp[cur][k][rem] = fact[rem + k][k];
        for (int j = 0; j < k; j++) dp[cur][j][rem] = 0;
      }
      for (int j = 1000; j > 0; j--) {
        cur = !cur;
        if (j > k)
          for (int rem = 0; rem <= 1000; rem++)
            dp[cur][k][rem] = fact[rem + k][k];
        for (int h = 0; h < k; h++) {
          if (h >= j) {
            break;
          }
          for (int rem = 1000; rem >= 0; rem--) {
            if ((k - h) * (j * 2 + k - h - 1) > 2 * rem) {
              dp[cur][h][rem] = 0;
              break;
            }
            dp[cur][h][rem] = dp[!cur][h + 1][rem - j];
            if ((k - h) * ((1 + j) * 2 + k - h - 1) <= 2 * rem)
              dp[cur][h][rem] += dp[!cur][h][rem];
            if (dp[cur][h][rem] >= (1000000007))
              dp[cur][h][rem] -= (1000000007);
          }
        }
      }
    }
    ans[v[i].second.second] = dp[cur][0][n];
  }
  for (int i = 0; i < t; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Your challenge is to write 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;
int dp[3][50][1002], fact[1050][50], ans[((int)(2e5 + 1))];
vector<pair<int, pair<int, int> > > v;
bool cmp(pair<int, pair<int, int> > a, pair<int, pair<int, int> > b) {
  if (a.first == b.first) return a.second.first > b.second.first;
  return a.first < b.first;
}
int main() {
  for (int i = 1; i <= 1046; i++) {
    fact[i][0] = 1;
    for (int j = 1; j <= min(i, 46); j++) {
      fact[i][j] = ((long long)fact[i][j - 1] * (i - j + 1)) % (1000000007);
    }
  }
  int t, n, k, tc;
  scanf("%d", &t);
  for (int i = 0; i < t; i++) {
    scanf("%d %d", &n, &k);
    v.push_back(make_pair(k, make_pair(n, i)));
  }
  sort(v.begin(), v.end(), cmp);
  bool cur;
  for (int i = 0; i < v.size(); i++) {
    k = v[i].first;
    n = v[i].second.first;
    if (k * (k + 1) > 2 * n) {
      ans[v[i].second.second] = 0;
      continue;
    }
    if (i == 0 || v[i].first != v[i - 1].first) {
      cur = 0;
      for (int rem = 0; rem <= 1000; rem++) {
        dp[cur][k][rem] = fact[rem + k][k];
        for (int j = 0; j < k; j++) dp[cur][j][rem] = 0;
      }
      for (int j = 1000; j > 0; j--) {
        cur = !cur;
        if (j > k)
          for (int rem = 0; rem <= 1000; rem++)
            dp[cur][k][rem] = fact[rem + k][k];
        for (int h = 0; h < k; h++) {
          if (h >= j) {
            break;
          }
          for (int rem = 1000; rem >= 0; rem--) {
            if ((k - h) * (j * 2 + k - h - 1) > 2 * rem) {
              dp[cur][h][rem] = 0;
              break;
            }
            dp[cur][h][rem] = dp[!cur][h + 1][rem - j];
            if ((k - h) * ((1 + j) * 2 + k - h - 1) <= 2 * rem)
              dp[cur][h][rem] += dp[!cur][h][rem];
            if (dp[cur][h][rem] >= (1000000007))
              dp[cur][h][rem] -= (1000000007);
          }
        }
      }
    }
    ans[v[i].second.second] = dp[cur][0][n];
  }
  for (int i = 0; i < t; i++) printf("%d\n", ans[i]);
  return 0;
}

```