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.ArrayList;
import java.util.HashMap;
import java.util.Scanner;

public class Main {
	
	public static void main(String[] args)
	{
		Scanner s=new Scanner(System.in);
		
		StringBuilder sb=new StringBuilder();
		
		int t=s.nextInt();
		
		for(int i=0;i<t;i++)
		{
			int n=s.nextInt();
			
			int[] arr=new int[n];
			
			for(int j=0;j<n;j++)
			{
				arr[j]=s.nextInt();
			}
			
			ArrayList<Integer> list=new ArrayList<>();
			int pre=0;
			
			int p=0;
			
			if(arr[1]-arr[0]>0)
			{
				p=0;
			}
			else
			{
				p=1;
			}
			
			for(int j=1;j<n;j++)
			{
				if(arr[j]>arr[j-1]&&p==0)
				{
					
				}
				else if(arr[j]>arr[j-1]&&p==1)
				{
					list.add(pre);
					list.add(j-1);
					
					pre=j-1;
					p=0;
				}
				else if(arr[j]<arr[j-1]&&p==1)
				{
					
				}
				else
				{
					list.add(pre);
					list.add(j-1);
					
					pre=j-1;
					
					p=1;
				}
			}
			
			list.add(pre);
			list.add(n-1);
			
			HashMap<Integer,Integer> map=new HashMap<>();
			
			ArrayList<Integer> temp=new ArrayList<>();
			
			int count=0;
			
			for(int j=0;j<list.size();j++)
			{
				if(!map.containsKey(list.get(j)))
				{
					count++;
					temp.add(list.get(j));
					map.put(list.get(j),1);
				}
			}
			
			sb.append(count);
			sb.append("\n");
			
			for(int j=0;j<temp.size();j++)
			{
				sb.append(arr[temp.get(j)]+" ");
			}
			
			sb.append("\n");
			
		}
		
		System.out.println(sb);
	}
	
}