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:
//package codeforces;

import java.util.Scanner;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.TreeMap;
import java.util.TreeSet;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.Set;

public class q {
	// Integer.bitCount(i); count no of 1 bit in a given number
	static Scanner scn = new Scanner(System.in);
	// static int min=Integer.MAX_VALUE;

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

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

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

	static FastReader s = new FastReader();

	static int mod = 1000000007;

	public static void main(String[] args) {
		// System.out.format("%.10f", ans);char c = sc.next().charAt(0);
		// code
		int t = scn.nextInt();
		while (t-- > 0) {
			int n=scn.nextInt();
			int arr[]=new int[n];
			for(int i=0;i<n;i++)
				arr[i]=scn.nextInt();
			ArrayList<Integer>list=new ArrayList<Integer>();
			list.add(arr[0]);
			for(int i=1;i<n;i++)
			{
				if(i<n-1)
				{
					if((arr[i]<arr[i+1]&&arr[i]>arr[i-1])||(arr[i]>arr[i+1]&&arr[i]<arr[i-1]))
					{
						//list.add(arr[i]);
					}
					else
						list.add(arr[i]);
				}
				else if(i==n-1)
				{
					list.add(arr[i]);
				}
			}
			long sum=0;
			for(int i=0;i<list.size()-1;i++)
			{
				sum+=Math.abs(list.get(i)-list.get(i+1));
			}
			System.out.println(list.size());
			for(int i=0;i<list.size();i++)
			{
				System.out.print(list.get(i)+" ");
			}
			System.out.println();
			//System.out.println(list);
				
			}
	

	}
	public static long cut(int n,int arr[],int prev,int sum,long memo[])
	{if(sum==n)
		return 0;
	if(sum>n)
		return Integer.MIN_VALUE;
	if(memo[sum]!=-1)
		return memo[sum];
		int l=arr.length;
		long a=Integer.MIN_VALUE,b=Integer.MIN_VALUE;
		for(int i=prev;i<l;i++)
		{
			b=cut(n, arr, i,sum+arr[i],memo)+1;
			a=Math.max(a, b);
			memo[sum]=a;
		}
		return a;
	}

	public static boolean[] sieveOfEratosthenes(int n) {

		boolean prime[] = new boolean[n + 1];
		for (int i = 0; i < n; i++)
			prime[i] = true;

		for (int p = 2; p * p <= n; p++) {
			// If prime[p] is not changed, then it is a prime
			if (prime[p] == true) {
				// Update all multiples of p
				for (int i = p * p; i <= n; i += p)
					prime[i] = false;
			}
		}

		for (int i = 2; i <= n; i++) {
			if (prime[i] == true) {
				// list.add(i);
			}

		}
		return prime;
	}

	public static class Pair implements Comparable {
		int x;
		int y;
		int z;

		Pair(int x, int y, int z) {
			this.x = x;
			this.y = y;
			this.z = z;
		}

		public int compareTo(Object o) {
			Pair pp = (Pair) o;
			if (pp.x == x)
				return 0;
			else if (x > pp.x)
				return 1;
			else
				return -1;
		}

	}

}