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;

/* package codechef; // don't place package name! */



import java.util.Arrays;
import java.util.Collections;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.TreeMap;



public class TP {
	
	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 a[]=new int[n]; 
			int sum=0;
			for(int i=0;i<n;i++)
			{
				a[i]=sc.nextInt();
			}
			boolean flag=false;
			if(a[1]-a[0]>0)
			{
				flag=true;
			}
			ArrayList<Integer> al=new ArrayList<>();
			al.add(a[0]);
			for(int i=2;i<n;i++)
			{
				if(flag==true && a[i]-a[i-1]<0)
				{
					flag=false;
					al.add(a[i-1]);
				}
				else if(flag==false && a[i]-a[i-1]>0)
				{
					flag=true;
					al.add(a[i-1]);
				}
			}
			al.add(a[n-1]);
			System.out.println(al.size());
			for(int i=0;i<al.size();i++)
			{
				System.out.print(al.get(i)+" ");
			}
			System.out.println();
		}
	}
}