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 B28
{ 
  public static void main(String[] args)
   {
    Scanner s=new Scanner(System.in);
    int t=s.nextInt();
    while(t>0)
     {
      int n=s.nextInt();
      int a[]=new int[n+1];
	
	int l=0;
	
	for(int i=1;i<=n;i++)
	 {
	 a[i]=s.nextInt();
	 }
	ArrayList<Integer> arr=new ArrayList<Integer>();
	boolean b[]=new boolean[n+1];
	int i=1;	
	
	arr.add(a[1]);
	b[a[1]]=true;
        for(int j=2;j<n;j++)
	{
	  //System.out.println(i+ " "+j);
	  if(a[j]>a[i]&&a[j+1]<a[j])
	   i=j;
	  else if(a[j]<a[i]&&a[j+1]>a[j])
	   i=j;
	  if(!b[a[i]])
	  {arr.add(a[i]);b[a[i]]=true;}
	 
	 
	}
	if(!b[a[n]])
        arr.add(a[n]);
	System.out.println(arr.size());
	Iterator<Integer> ir=arr.listIterator();
	while(ir.hasNext())
	{
	 System.out.print(ir.next()+" ");
	}
	
	System.out.println();
      t--;
     }
   }
}