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.util.*;
import java.io.*;

public class Main {
    public static void main(String args[]) {new Main().run();}

    FastReader in = new FastReader();
    PrintWriter out = new PrintWriter(System.out);
    void run() {
        for(int q=ni();q>0;q--){
            work();
        }
        out.flush();
    }
    long mod=1000000007;
    long gcd(long a,long b) {
        return a==0?b:gcd(b%a,a);
    }
    void work() {
        int n=ni(),k=ni();
        int[][] dp=new int[n+1][n+1];
        int[] A=new int[n+1];
        for(int i=1;i<=n;i++){
            A[i]=ni();
        }
        for(int i=1;i<=n;i++){
            for(int j=0;j<=i;j++){
                dp[i][j]=dp[i-1][j]+(A[i]==i-j?1:0);
                if(j>0){
                    dp[i][j]=Math.max(dp[i][j],dp[i-1][j-1]);
                }
            }
        }
        for(int j=0;j<=n;j++){
            if(dp[n][j]>=k){
                out.println(j);
                return;
            }
        }
        out.println(-1);
    }

    //input
    @SuppressWarnings("unused")
    private ArrayList<Integer>[] ng(int n, int m) {
        ArrayList<Integer>[] graph=(ArrayList<Integer>[])new ArrayList[n];
        for(int i=0;i<n;i++) {
            graph[i]=new ArrayList<>();
        }
        for(int i=1;i<=m;i++) {
            int s=in.nextInt()-1,e=in.nextInt()-1;
            graph[s].add(e);
            graph[e].add(s);
        }
        return graph;
    }

    private ArrayList<long[]>[] ngw(int n, int m) {
        ArrayList<long[]>[] graph=(ArrayList<long[]>[])new ArrayList[n];
        for(int i=0;i<n;i++) {
            graph[i]=new ArrayList<>();
        }
        for(int i=1;i<=m;i++) {
            long s=in.nextLong()-1,e=in.nextLong()-1,w=in.nextLong();
            graph[(int)s].add(new long[] {e,w});
            graph[(int)e].add(new long[] {s,w});
        }
        return graph;
    }

    private int ni() {
        return in.nextInt();
    }

    private long nl() {
        return in.nextLong();
    }
    private double nd() {
        return in.nextDouble();
    }
    private String ns() {
        return in.next();
    }

    private long[] na(int n) {
        long[] A=new long[n];
        for(int i=0;i<n;i++) {
            A[i]=in.nextLong();
        }
        return A;
    }

    private int[] nia(int n) {
        int[] A=new int[n];
        for(int i=0;i<n;i++) {
            A[i]=in.nextInt();
        }
        return A;
    }
}

class FastReader
{
    BufferedReader br;
    StringTokenizer st;
    InputStreamReader input;//no buffer
    public FastReader()
    {
        br=new BufferedReader(new InputStreamReader(System.in));
    }

    public FastReader(boolean isBuffer)
    {
        if(!isBuffer){
            input=new InputStreamReader(System.in);
        }else{
            br=new BufferedReader(new InputStreamReader(System.in));
        }
    }

    public boolean hasNext(){
        try{
            String s=br.readLine();
            if(s==null){
                return  false;
            }
            st=new StringTokenizer(s);
        }catch(IOException e){
            e.printStackTrace();
        }
        return true;
    }

    public String next()
    {   
        if(input!=null){
            try {
                StringBuilder sb=new StringBuilder();
                int ch=input.read();
                while(ch=='\n'||ch=='\r'||ch==32){
                    ch=input.read();
                }
                while(ch!='\n'&&ch!='\r'&&ch!=32){
                    sb.append((char)ch);
                    ch=input.read();
                }
                return sb.toString();
            }catch (Exception e){
                e.printStackTrace();
            }
        }
        while(st==null || !st.hasMoreElements())//回车，空行情况
        {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    public int nextInt()
    {
        return (int)nextLong();
    }

    public long nextLong() {
        try {
            if(input!=null){
                long ret=0;
                int b=input.read();
                while(b<'0'||b>'9'){
                    b=input.read();
                }
                while(b>='0'&&b<='9'){
                    ret=ret*10+(b-'0');
                    b=input.read();
                }
                return ret;
            }
        }catch (Exception e){
            e.printStackTrace();
        }
        return Long.parseLong(next());
    }

    public double nextDouble()
    {
        return Double.parseDouble(next());
    }
}