To help those contestants who struggle a lot in contests, the headquarters of Codeforces are planning to introduce Division 5. In this new division, the tags of all problems will be announced prior to the round to help the contestants.

The contest consists of n problems, where the tag of the i-th problem is denoted by an integer a_i.

You want to AK (solve all problems). To do that, you must solve the problems in some order. To make the contest funnier, you created extra limitations on yourself. You do not want to solve two problems consecutively with the same tag since it is boring. Also, you are afraid of big jumps in difficulties while solving them, so you want to minimize the number of times that you solve two problems consecutively that are not adjacent in the contest order.

Formally, your solve order can be described by a permutation p of length n. The cost of a permutation is defined as the number of indices i (1≤ i<n) where |p_{i+1}-p_i|>1. You have the requirement that a_{p_i}≠ a_{p_{i+1}} for all 1≤ i< n.

You want to know the minimum possible cost of permutation that satisfies the requirement. If no permutations meet this requirement, you should report about it.

Input

The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.

The first line of the description of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of problems in the contest.

The next line contains n integers a_1,a_2,… a_n (1 ≤ a_i ≤ n) — the tags of the problems.

It is guaranteed that the sum of n over all test cases does not exceed 10^5.

Output

For each test case, if there are no permutations that satisfy the required condition, print -1. Otherwise, print the minimum possible cost of a permutation that satisfies the required condition.

Example

Input


4
6
2 1 2 3 1 1
5
1 1 1 2 2
8
7 7 2 7 7 1 8 7
10
1 2 3 4 1 1 2 3 4 1


Output


1
3
-1
2

Note

In the first test case, let p=[5, 4, 3, 2, 1, 6]. The cost is 1 because we jump from p_5=1 to p_6=6, and |6-1|>1. This permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 1.

In the second test case, let p=[1,5,2,4,3]. The cost is 3 because |p_2-p_1|>1, |p_3-p_2|>1, and |p_4-p_3|>1. The permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 3.

In the third test case, for any order of solving the problems, we will solve two problems with the same tag consecutively, so the answer is -1.
import java.io.*;
import java.util.InputMismatchException;
import java.util.function.IntUnaryOperator;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
	static class TaskAdapter implements Runnable {
		@Override
		public void run() {
			long startTime = System.currentTimeMillis();
			InputStream inputStream = System.in;
			OutputStream outputStream = System.out;
			FastReader in = new FastReader(inputStream);
			Output out = new Output(outputStream);
			FTheStrugglingContestant solver = new FTheStrugglingContestant();
			int testCount = in.nextInt();
			for(int i = 1; i<=testCount; i++)
				solver.solve(i, in, out);
			out.close();
			System.err.println(System.currentTimeMillis()-startTime+"ms");
		}
	}

	public static void main(String[] args) throws Exception {
		Thread thread = new Thread(null, new TaskAdapter(), "", 1<<24);
		thread.start();
		thread.join();
	}

	static class FTheStrugglingContestant {
		public FTheStrugglingContestant() {
		}

		public void solve(int kase, InputReader in, Output pw) {
			int n = in.nextInt();
			int[] arr = in.nextInt(n, o -> o-1);
			int max = 0, mcnt = 0, x = 0;
			int[] adj = new int[n], cnt = new int[n];
			adj[arr[0]]++;
			adj[arr[n-1]]++;
			cnt[arr[n-1]]++;
			for(int i = 0; i<n-1; i++) {
				if(arr[i]==arr[i+1]) {
					max = Math.max(max, adj[arr[i]] += 2);
					x++;
				}
				mcnt = Math.max(mcnt, ++cnt[arr[i]]);
			}
			if(mcnt>(n+1)/2) {
				pw.println("-1");
				return;
			}
			pw.println(x+Math.max(0, max-(x+2)));
		}

	}

	static class FastReader implements InputReader {
		final private int BUFFER_SIZE = 1<<16;
		private final DataInputStream din;
		private final byte[] buffer;
		private int bufferPointer;
		private int bytesRead;

		public FastReader(InputStream is) {
			din = new DataInputStream(is);
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public int nextInt() {
			int ret = 0;
			byte c = skipToDigit();
			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;
		}

		private boolean isDigit(byte b) {
			return b>='0'&&b<='9';
		}

		private byte skipToDigit() {
			byte ret;
			while(!isDigit(ret = read())&&ret!='-') ;
			return ret;
		}

		private void fillBuffer() {
			try {
				bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
			}catch(IOException e) {
				e.printStackTrace();
				throw new InputMismatchException();
			}
			if(bytesRead==-1) {
				buffer[0] = -1;
			}
		}

		private byte read() {
			if(bytesRead==-1) {
				throw new InputMismatchException();
			}else if(bufferPointer==bytesRead) {
				fillBuffer();
			}
			return buffer[bufferPointer++];
		}

	}

	interface InputReader {
		int nextInt();

		default int[] nextInt(int n, IntUnaryOperator operator) {
			int[] ret = new int[n];
			for(int i = 0; i<n; i++) {
				ret[i] = operator.applyAsInt(nextInt());
			}
			return ret;
		}

	}

	static class Output implements Closeable, Flushable {
		public StringBuilder sb;
		public OutputStream os;
		public int BUFFER_SIZE;
		public String lineSeparator;

		public Output(OutputStream os) {
			this(os, 1<<16);
		}

		public Output(OutputStream os, int bs) {
			BUFFER_SIZE = bs;
			sb = new StringBuilder(BUFFER_SIZE);
			this.os = new BufferedOutputStream(os, 1<<17);
			lineSeparator = System.lineSeparator();
		}

		public void println(int i) {
			println(String.valueOf(i));
		}

		public void println(String s) {
			sb.append(s);
			println();
		}

		public void println() {
			sb.append(lineSeparator);
		}

		private void flushToBuffer() {
			try {
				os.write(sb.toString().getBytes());
			}catch(IOException e) {
				e.printStackTrace();
			}
			sb = new StringBuilder(BUFFER_SIZE);
		}

		public void flush() {
			try {
				flushToBuffer();
				os.flush();
			}catch(IOException e) {
				e.printStackTrace();
			}
		}

		public void close() {
			flush();
			try {
				os.close();
			}catch(IOException e) {
				e.printStackTrace();
			}
		}

	}
}

