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 b {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		
		while(t-->0)
		{
			int n = sc.nextInt();
			int[] s = new int[n];
			
			for(int i = 0; i < n  ;i++)
			{
				s[i] = sc.nextInt();
			}
			
			if(n == 2)
			{
				System.out.println(2);
				System.out.println(s[0]+" "+s[1]);
				continue;
			}
			boolean[] in = new boolean[n];
			//System.out.print(s[0]);
			ArrayList<Integer> ans = new ArrayList<>();
			
			ans.add(s[0]);
			for(int i = 0 ; i < n-2; i++)
			{
				boolean now;
				boolean next;
				if(s[i] > s[i+1])
				{
					now = true;
				}	
				else
					now = false;
				
				if(s[i+1] > s[i+2])
					next = true;
				else
					next = false;
				
				if(next != now)
					ans.add(s[i+1]);
			}
			
//			for(int i = 0; i < ans.size(); i++)
//				System.out.println("tese"+ans.get(i));
			ans.add(s[n-1]);
			
			int len = ans.size();
			//System.out.println(len);
			if(len == 2)
			{
				
			}
			else if((ans.get(len-1) > ans.get(len-2) && ans.get(len-2)>ans.get(len-3))||(ans.get(len-1) < ans.get(len-2) && ans.get(len-2)<ans.get(len-3)))
				ans.remove(len-2);

			
			System.out.println(ans.size());
			for(int i = 0; i < ans.size(); i++)
			{
				
				System.out.print(ans.get(i)+" ");
			}
			
			System.out.println();
				
			
			
		}
	}

}