#include <bits/stdc++.h>
std::mt19937 rng(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
const int MOD = 1e9 + 7;
const int ms = 2020;
const int mn = ms / 2;
const int mk = 65;
inline void add(int &a, int &b) { a = (a + b >= MOD ? a + b - MOD : a + b); }
long long fat[ms];
int table[mk][mn];
int comb[ms][ms];
void pre() {
  fat[0] = 1;
  for (int i = 1; i < ms; i++) {
    fat[i] = (long long)fat[i - 1] * i % MOD;
  }
  table[0][0] = 1;
  for (int v = 1; v < mn; v++) {
    for (int k = mk - 1; k > 0; k--) {
      for (int i = 0; i + v < mn; i++) {
        add(table[k][i + v], table[k - 1][i]);
      }
    }
  }
  for (int i = 0; i < ms; i++) {
    comb[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      comb[i][j] = (comb[i - 1][j] + comb[i - 1][j - 1]) % MOD;
    }
  }
}
long long sum(int n, int k) { return (long long)comb[n + k - 1][k - 1]; }
int solve(int n, int k) {
  long long ans = 0;
  for (int i = k * (k + 1) / 2; i <= n; i++) {
    if (!table[k][i]) continue;
    long long cur = table[k][i];
    cur = cur * sum(n - i, k + 1) % MOD;
    ans = (ans + cur) % MOD;
  }
  ans = ans * fat[k] % MOD;
  return (int)ans;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  pre();
  int t;
  std::cin >> t;
  while (t--) {
    int n, k;
    std::cin >> n >> k;
    std::cout << solve(n, k) << std::endl;
  }
}

### 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>
std::mt19937 rng(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
const int MOD = 1e9 + 7;
const int ms = 2020;
const int mn = ms / 2;
const int mk = 65;
inline void add(int &a, int &b) { a = (a + b >= MOD ? a + b - MOD : a + b); }
long long fat[ms];
int table[mk][mn];
int comb[ms][ms];
void pre() {
  fat[0] = 1;
  for (int i = 1; i < ms; i++) {
    fat[i] = (long long)fat[i - 1] * i % MOD;
  }
  table[0][0] = 1;
  for (int v = 1; v < mn; v++) {
    for (int k = mk - 1; k > 0; k--) {
      for (int i = 0; i + v < mn; i++) {
        add(table[k][i + v], table[k - 1][i]);
      }
    }
  }
  for (int i = 0; i < ms; i++) {
    comb[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      comb[i][j] = (comb[i - 1][j] + comb[i - 1][j - 1]) % MOD;
    }
  }
}
long long sum(int n, int k) { return (long long)comb[n + k - 1][k - 1]; }
int solve(int n, int k) {
  long long ans = 0;
  for (int i = k * (k + 1) / 2; i <= n; i++) {
    if (!table[k][i]) continue;
    long long cur = table[k][i];
    cur = cur * sum(n - i, k + 1) % MOD;
    ans = (ans + cur) % MOD;
  }
  ans = ans * fat[k] % MOD;
  return (int)ans;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  pre();
  int t;
  std::cin >> t;
  while (t--) {
    int n, k;
    std::cin >> n >> k;
    std::cout << solve(n, k) << std::endl;
  }
}

```