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 _1551_E {
	public static void main(String[] args) throws IOException {
		Reader in = new Reader();
		PrintWriter out = new PrintWriter(System.out);
		int t = in.nextInt();
		while(t-- > 0) {
            int n = in.nextInt();
            int k = in.nextInt();
            int[] a = new int[n];
            for(int i = 0; i < n; i++) {
                a[i] = in.nextInt();
            }
            int max = -1;
            //dp[i][j] stores max fixed points using j elements from 0 to i - 1
            int[][] dp = new int[n + 1][n + 1];
            for(int i = 1; i <= n; i++) {
                for(int j = 1; j <= i; j++) {
                    if(i - 1 >= j) {
                        dp[i][j] = dp[i - 1][j];
                    }
                    dp[i][j] = Math.max(dp[i][j], dp[i - 1][j - 1] + (a[i - 1] == j ? 1 : 0));
                    if(dp[i][j] >= k) {
                        max = Math.max(max, j);
                    }
                }
            }
            if(max == -1) {
                out.println(max);
            }else {
                out.println(n - max);
            }
		}
		in.close();
		out.close();
	}
	static class Reader {
		final private int BUFFER_SIZE = 1 << 16;
		private DataInputStream din;
		private byte[] buffer;
		private int bufferPointer, bytesRead;
		public Reader() {
			din = new DataInputStream(System.in);
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}
		public Reader(String file_name) throws IOException {
			din = new DataInputStream(new FileInputStream(file_name));
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}
		public String readLine() throws IOException {
			byte[] buf = new byte[64]; // line length
			int cnt = 0, c;
			while((c = read()) != -1) {
				if(c == '\n') {
					if(cnt != 0) {
						break;
					}else {
						continue;
					}
				}
				buf[cnt++] = (byte) c;
			}
			return new String(buf, 0, cnt);
		}
		public int nextInt() throws IOException {
			int ret = 0;
			byte c = read();
			while(c <= ' ') {
				c = read();
			}
			boolean neg = (c == '-');
			if(neg) {
				c = read();
			}
			do{
				ret = ret * 10 + c - '0';
			}while((c = read()) >= '0' && c <= '9');
			if(neg) {
				return -ret;
			}
			return ret;
		}
		public long nextLong() throws IOException {
			long ret = 0;
			byte c = read();
			while(c <= ' ') {
				c = read();
			}
			boolean neg = (c == '-');
			if(neg) {
				c = read();
			}
			do{
				ret = ret * 10 + c - '0';
			}while((c = read()) >= '0' && c <= '9');
			if(neg) {
				return -ret;
			}
			return ret;
		}
		public double nextDouble() throws IOException {
			double ret = 0, div = 1;
			byte c = read();
			while(c <= ' ') {
				c = read();
			}
			boolean neg = (c == '-');
			if(neg) {
				c = read();
			}
			do{
				ret = ret * 10 + c - '0';
			}while((c = read()) >= '0' && c <= '9');
			if(c == '.') {
				while((c = read()) >= '0' && c <= '9') {
					ret += (c - '0') / (div *= 10);
				}
			}
			if(neg) {
				return -ret;
			}
			return ret;
		}
		private void fillBuffer() throws IOException {
			bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
			if(bytesRead == -1) {
				buffer[0] = -1;
			}
		}
		private byte read() throws IOException {
			if(bufferPointer == bytesRead) {
				fillBuffer();
			}
			return buffer[bufferPointer++];
		}
		public void close() throws IOException {
			if(din == null) {
				return;
			}
			din.close();
		}
	}
}
