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:
/*input
5 5 2 3



*/
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
 {
 static Scanner in = new Scanner(System.in);
 
 
    public static void solve()
    {   
        int n=in.nextInt();
        int[] a = new int[n];
        for(int i=0;i<n;i++)
        {
            a[i]=in.nextInt();
        }
        HashSet<Integer> hs = new HashSet<>();
        
      for(int i=1;i<n-1;i++)
      {
          if( (a[i-1]<a[i])==(a[i]<a[i+1]))
          {
              hs.add(i);
             
              
          }
         
      }
      StringBuilder sb = new StringBuilder();
      for(int i=0;i<n;i++) 
      {
        if(!hs.contains(i))
        {
            sb.append(a[i]).append(" ");
        }
      }
      System.out.println(n-hs.size());
      System.out.println(sb);
    	
    }
    
	public static void main (String[] args)
	 {
	 int t=in.nextInt();
	 while(t-->0)
	 {
	  solve();
	 }
}
}