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.

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public final class soltion{
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        int count= Integer.parseInt(bufferedReader.readLine());
    StringBuilder stringBuilder= new StringBuilder();
    for (int t = 0; t < count; t++) {
        String[] input= bufferedReader.readLine().split(" ");
        int size=Integer.parseInt(input[0]);
        int minEqual=Integer.parseInt(input[1]);
        String[] in= bufferedReader.readLine().split(" ");
        int[] numbers=new int[size];
        int ans=-1;
        for (int i = 0; i < size; i++) {
            numbers[i]=Integer.parseInt(in[i]);
        }
       int [][] dp= new int[size][size+1];
        if(numbers[0]==1)dp[0][0]=1;
        for (int i = 1; i < size; i++) {
            dp[i][0]=dp[i-1][0];
            if(numbers[i]==i+1){
                dp[i][0]++;
            }
        }
        for (int j = 1; j <= size; j++) {
            boolean tobebroken=false;
            for (int i = j; i <size; i++) {
                if(numbers[i]==i-j+1){
                    dp[i][j]=1+dp[i-1][j];
                    if(dp[i][j]==minEqual){
                        ans=j;
                        tobebroken=true;
                        break;
                    }
                }else {
                    dp[i][j]=Math.max(dp[i-1][j-1],dp[i-1][j]);
                }
            }
            if(tobebroken)break;
            
        }
        stringBuilder.append(ans);
        stringBuilder.append("\n");
    }
    System.out.println(stringBuilder.toString());
    
}
}