Problem: Given a permutation p of length n, find its subsequence s_1, s_2, …, s_k of length at least 2 such that:

  * |s_1-s_2|+|s_2-s_3|+…+|s_{k-1}-s_k| is as big as possible over all subsequences of p with length at least 2. 
  * Among all such subsequences, choose the one whose length, k, is as small as possible. 



If multiple subsequences satisfy these conditions, you are allowed to find any of them.

A sequence a is a subsequence of an array b if a can be obtained from b by deleting some (possibly, zero or all) elements.

A permutation of length n is an array of length n in which every element from 1 to n occurs exactly once.

Input

The first line contains an integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer n (2 ≤ n ≤ 10^5) — the length of the permutation p.

The second line of each test case contains n integers p_1, p_2, …, p_{n} (1 ≤ p_i ≤ n, p_i are distinct) — the elements of the permutation p.

The sum of n across the test cases doesn't exceed 10^5.

Output

For each test case, the first line should contain the length of the found subsequence, k. The second line should contain s_1, s_2, …, s_k — its elements.

If multiple subsequences satisfy these conditions, you are allowed to find any of them.

Example

Input


2
3
3 2 1
4
1 3 4 2


Output


2
3 1 
3
1 4 2 

Note

In the first test case, there are 4 subsequences of length at least 2:

  * [3,2] which gives us |3-2|=1. 
  * [3,1] which gives us |3-1|=2. 
  * [2,1] which gives us |2-1|=1. 
  * [3,2,1] which gives us |3-2|+|2-1|=2. 



So the answer is either [3,1] or [3,2,1]. Since we want the subsequence to be as short as possible, the answer is [3,1].

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
import java.math.*;
 
public class Main {

	private static FastReader fr = new FastReader();
	private static Helper helper = new Helper();
	private static StringBuilder result = new StringBuilder();
 
	public static void main(String[] args) {
		Task solver = new Task();
		solver.solve();
	}
 
	static class Task {
	    
		public void solve() {
		    int nt = fr.ni();
			while(nt-- > 0){
				int n = fr.ni();
				Long[] arr = new Long[n];
				for(int i=0; i<n; i++) arr[i] = fr.nl();
				ArrayList<Long> resultList = new ArrayList<>();
				for(int i=0; i<n; i++){
					if(i == 0 || i == n-1){
						resultList.add(arr[i]);
					}
					else{
						if((arr[i] > arr[i-1] && arr[i] > arr[i+1]) || 
								(arr[i] < arr[i+1] && arr[i] < arr[i-1])) resultList.add(arr[i]);
					}
				}
				result.append(resultList.size()).append("\n");
				for(long val : resultList) result.append(val).append(" ");
				result.append("\n");
			}
			System.out.println(result);
		}
	}
    
    static class Helper{
		public long[] tiArr(int n, int si){
			long[] arr = new long[n];
			for(int i=si; i<n; i++) arr[i] = fr.nl();
			return arr;
		}
	}
 
	static class FastReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;
		private static PrintWriter pw;
 
		public FastReader() {
			reader = new BufferedReader(new InputStreamReader(System.in));
			pw = new PrintWriter(System.out);
			tokenizer = null;
		}
 
		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
 
		public int ni() {
			return Integer.parseInt(next());
		}
 
		public long nl() {
			return Long.parseLong(next());
		}
 
		public String rl() {
			try {
				return reader.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return null;
		}
 
		public void print(String str) {
			pw.print(str);
			pw.flush();
		}
	}
}