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.*;
import java.util.*;
import java.math.BigInteger;
import java.util.InputMismatchException;

public class Main {
	static PrintWriter out;
	static Reader in;
	public static void main(String[] args) throws IOException {
		input_output();
		Main solver = new Main();
		solver.solve();
		out.close();

		out.flush();
	} 

	static long INF = (long)1e18;
	static int MAXN = (int)1e5 + 5;
	static int MOD = (int)1e9 + 7;
	static int q, t, n, m, k;
	static double pi = Math.PI;

	void solve() throws IOException {
		t = in.nextInt();
		

		while (t --> 0) {
			n = in.nextInt();
			k = in.nextInt();

			int[] arr = new int[n+1];
			for (int i = 1; i <= n; i++) arr[i] = in.nextInt();

			int[][] dp = new int[n+1][n+2];

			for (int i = 1; i <= n; i++) {
				for (int j = 0; j <= n; j++) {
					dp[i][j] = Math.max(dp[i][j], dp[i-1][j]+(j == i-arr[i] ? 1: 0));
					dp[i][j+1] = Math.max(dp[i][j+1], dp[i-1][j]);
				}

			}


			int ans = -1;
			for (int i = 0; i <= n; i++) {
				if (dp[n][i] >= k) {
					ans = i;
					break;
				}
			}

			out.println(ans);
		}	
	}

	static class Reader {
		private InputStream mIs;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;

		public Reader() {
			this(System.in);
		}

		public Reader(InputStream is) {
			mIs = is;
		}

		public int read() {
			if (numChars == -1) {
				throw new InputMismatchException();

			}
			if (curChar >= numChars) {
				curChar = 0;
				try {
					numChars = mIs.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (numChars <= 0) {
					return -1;
				}
			}
			return buf[curChar++];
		}

		public String nextLine() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndOfLine(c));
			return res.toString();
		}

		public String next() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

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

		public long nextLong() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long res = 0;
			do {
				if (c < '0' || c > '9') {
					throw new InputMismatchException();
				}
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public int nextInt() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9') {
					throw new InputMismatchException();
				}
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		public boolean isEndOfLine(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}

	}
	static void input_output() throws IOException {
		File f = new File("in.txt");
		if (f.exists() && !f.isDirectory()) {
			in = new Reader(new FileInputStream("in.txt"));
		} else in = new Reader();
		f = new File("out.txt");
		if (f.exists() && !f.isDirectory()) {
			out = new PrintWriter(new File("out.txt"));
		} else out = new PrintWriter(System.out);
	}
}