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.*;
public class Mostsociallydistancedsubsequence {
	public static void main(String[] args)
	{
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0)
		{
			int n=sc.nextInt();
			int arr[]=new int[n];
			boolean inc[]=new boolean[n];
			for(int i=0;i<n;i++)
			{
				arr[i]=sc.nextInt();
			}
			inc[0]=true;
			for(int i=1;i<n;i++)
			{
				if(arr[i]>arr[i-1])
				{
				inc[i]=true;	
				}else {
					inc[i]=false;
				}
			}
			int f=0;
			
			ArrayList<Integer> ans=new ArrayList<Integer>();
			ans.add(arr[f]);
			for(int i=2;i<n;i++)
			{
				if((inc[i]&&inc[i-1])||(!inc[i]&&!inc[i-1]))
				{
					continue;
				}else {
					
					
					f=i-1;
					ans.add(arr[f]);
				}
			}
			ans.add(arr[n-1]);
			 System.out.println(ans.size());
			for(int i=0;i<ans.size();i++)
		  {
			 
			  System.out.print(ans.get(i)+" ");
		  }
		}
	}
}