#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n, k;
  cin >> n >> k;
  vector<vector<int>> dp(n + 1, vector<int>(n + 1, 0));
  vector<int> arr(n + 1, 0);
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
    for (int j = i; j >= 1; j--) {
      dp[i][j] = max(dp[i][j], dp[i - 1][j]);
      dp[i][j] =
          max(dp[i][j], arr[i] == j ? 1 + dp[i - 1][j - 1] : dp[i - 1][j - 1]);
    }
  }
  for (int j = n; j >= 0; j--) {
    if (dp[n][j] >= k) {
      cout << n - j << endl;
      return;
    }
  }
  cout << -1 << endl;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

In Cpp, your task is to solve the following problem:

Consider a sequence of integers a_1, a_2, …, a_n. In one move, you can select any element of the sequence and delete it. After an element is deleted, all elements to the right are shifted to the left by 1 position, so there are no empty spaces in the sequence. So after you make a move, the sequence's length decreases by 1. The indices of the elements after the move are recalculated.

E. g. let the sequence be a=[3, 2, 2, 1, 5]. Let's select the element a_3=2 in a move. Then after the move the sequence will be equal to a=[3, 2, 1, 5], so the 3-rd element of the new sequence will be a_3=1 and the 4-th element will be a_4=5.

You are given a sequence a_1, a_2, …, a_n and a number k. You need to find the minimum number of moves you have to make so that in the resulting sequence there will be at least k elements that are equal to their indices, i. e. the resulting sequence b_1, b_2, …, b_m will contain at least k indices i such that b_i = i.

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.

Each test case consists of two consecutive lines. The first line contains two integers n and k (1 ≤ k ≤ n ≤ 2000). The second line contains a sequence of integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n). The numbers in the sequence are not necessarily different.

It is guaranteed that the sum of n over all test cases doesn't exceed 2000.

Output

For each test case output in a single line:

  * -1 if there's no desired move sequence; 
  * otherwise, the integer x (0 ≤ x ≤ n) — the minimum number of the moves to be made so that the resulting sequence will contain at least k elements that are equal to their indices. 

Example

Input


4
7 6
1 1 2 3 4 5 6
5 2
5 1 3 2 3
5 2
5 5 5 5 4
8 4
1 2 3 3 2 2 5 5


Output


1
2
-1
2

Note

In the first test case the sequence doesn't satisfy the desired condition, but it can be provided by deleting the first element, hence the sequence will be [1, 2, 3, 4, 5, 6] and 6 elements will be equal to their indices.

In the second test case there are two ways to get the desired result in 2 moves: the first one is to delete the 1-st and the 3-rd elements so that the sequence will be [1, 2, 3] and have 3 elements equal to their indices; the second way is to delete the 2-nd and the 3-rd elements to get the sequence [5, 2, 3] with 2 desired elements.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n, k;
  cin >> n >> k;
  vector<vector<int>> dp(n + 1, vector<int>(n + 1, 0));
  vector<int> arr(n + 1, 0);
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
    for (int j = i; j >= 1; j--) {
      dp[i][j] = max(dp[i][j], dp[i - 1][j]);
      dp[i][j] =
          max(dp[i][j], arr[i] == j ? 1 + dp[i - 1][j - 1] : dp[i - 1][j - 1]);
    }
  }
  for (int j = n; j >= 0; j--) {
    if (dp[n][j] >= k) {
      cout << n - j << endl;
      return;
    }
  }
  cout << -1 << endl;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```