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 cf649B {
    public static void main(String [] args){
        Scanner sc = new Scanner(System.in);
        int t=sc.nextInt();
        while(t>0){

            int n=sc.nextInt();
            ArrayList<Integer> arr=new ArrayList<Integer>();
            ArrayList<Integer> ans=new ArrayList<Integer>();

            for(int i=0;i<n;i++){
                arr.add(sc.nextInt());
            }

            int i1=0;
            while(i1<n){
                boolean dec=true;
                boolean inc=true;
                int x1=arr.get(i1);
                ans.add(x1);
                for(int j=i1+1;j<n;j++){
                    if(x1<=arr.get(j) && arr.get(j-1)<=arr.get(j) && inc==true){
                        inc=true;
                        dec=false;
                        i1=j;
                    }
                    else if(x1>=arr.get(j) && arr.get(j-1)>=arr.get(j) && dec==true){
                        dec=true;
                        inc=false;
                        i1=j;
                    }
                    else{
                        i1=j-1;
                        break;
                    }
                }
                if(i1==n-1) {
                    ans.add(arr.get(i1));
                    i1++;
                }
            }
            System.out.println(ans.size());
            for(int i=0;i<ans.size();i++)
                System.out.print(ans.get(i)+ " ");
            System.out.print('\n');
            t--;
        }
    }
}