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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.Iterator;
import java.util.StringTokenizer;

public class MostSociallyDistancedSubsequence {
	static class FastScanner 
	{ 
		BufferedReader br; 
		StringTokenizer st; 

		public FastScanner() 
		{ 
			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; 
		} 
	} 

	public static void main(String[] args) {
		FastScanner f=new FastScanner();
		int t=f.nextInt();
		while(t-->0) {
			int n=f.nextInt();
			int arr[]=new int[n];
			boolean visited[]=new boolean[n+1];
			int min=Integer.MAX_VALUE;
			int max=Integer.MIN_VALUE;
			int indmax=-1;
			int indmin=-1;
			for(int i=0;i<n;i++) {
				arr[i]=f.nextInt();
				if(arr[i]>max) {
					max=arr[i];
					indmax=i;
				}
				if(arr[i]<min) {
					min=arr[i];
					indmin=i;
				}
			}
			
			int ans=2;
			ArrayList<Integer> a=new ArrayList<Integer>();
			a.add(arr[0]);
			for(int i=1;i<n-1;i++) {
				if(arr[i]>arr[i+1]&&arr[i]>arr[i-1]) {
					ans++;
					a.add(arr[i]);
				}else if(arr[i]<arr[i+1]&&arr[i]<arr[i-1]) {
					ans++;
					a.add(arr[i]);
				}
			}
			a.add(arr[n-1]);
			System.out.println(ans);
			for(int x:a) {
				System.out.print(x+" ");
			}
			System.out.println();
			
		}

	}

}