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.*;
import java.lang.*;
import java.io.*;
 
/* Name of the class has to be "Main" only if the class is public. */
public class Codechef{
	public static void main (String[] args)throws IOException {
    try{
	   //BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
	   Scanner sc = new Scanner(System.in);
	//int t=Integer.parseInt(br.readLine());
	//int t=1;
	 int t=sc.nextInt();
	   while(t-->0){
	 //int n=Integer.parseInt(br.readLine())
      int n = sc.nextInt();
    //int m = sc.nextInt();
    //int k = sc.nextInt();
	//long d=Long.parseLong(br.readLine());
	//String[] s= br.readLine().trim().split(" ");
	 Map<Integer,Integer> m = new TreeMap<Integer,Integer>();
    //List<Character> f = new ArrayList<Character>();
    List<Integer> s = new ArrayList<Integer>();
    //List<Integer> f = new ArrayList<Integer>();
     //<Integer> v= new Vector<Integer>();
	 //Set<Integer> s= new HashSet<Integer>();
	int c;
	for(int i=0;i<n;i++){
	     c= sc.nextInt();
	    s.add(c);
	}
	for(int i=1;i<n-1;i++){
	    if(s.get(i)>s.get(i-1)&&s.get(i)>s.get(i+1) || s.get(i)<s.get(i-1)&&s.get(i)<s.get(i+1)) continue;
	    else m.put(i,0);
	}

        System.out.println(s.size()-m.size());
        for(int i=0;i<n;i++){
            if(!m.containsKey(i)) System.out.print(s.get(i)+" ");
        }
    System.out.println();
}	
}
catch(Exception e){
    
}
	}
	}

 class pair{
    int x,y,ind;
    pair(int n,int m,int id){
        x=n;
        y=m;
        ind=id;
    }
}
class comp1 implements Comparator<pair>{
    public int compare(pair p1,pair p2){
        if(p1.x>p2.x)return -1;
        else return 1;
    }
}

class comp2 implements Comparator<pair>{
    public int compare(pair p1,pair p2){
        if(p1.y<p2.y)return -1;
        else return 1;
    }
}