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.io.BufferedReader;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.*;
import java.io.*; 
import java.util.HashSet;

import javax.lang.model.util.ElementScanner6;

  
public class STRING{
     
      public static void main( String args[]) {  
         
      BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); 
       Scanner sc = new Scanner(System.in);  
      PrintWriter writer = new PrintWriter(System.out);
      int t = sc.nextInt();
      while(t-->0){
          long n = sc.nextLong();
          long arr[] = new long[(int)n];
          for(int i =0;i<n;i++)
            arr[i] = sc.nextLong();
           ArrayList<Long> arrli = new ArrayList<Long>();
           arrli.add(arr[0]); 
          for(int i =1;i<n-1;i++){
              if((arr[i]-arr[i-1])*(arr[i]-arr[i+1])>0){
                    arrli.add(arr[i]);
              }
              
           }
           arrli.add(arr[(int)n-1]);
           writer.println(arrli.size());
           for(int i=0;i<arrli.size();i++)
             writer.print(arrli.get(i)+" ");
             writer.println();
             writer.flush();

           
        
          
        
      }
      
      
        
}  

   public static String replace(String str, int index, char replace){     
    if(str==null){
        return str;
    }else if(index<0 || index>=str.length()){
        return str;
    }
    char[] chars = str.toCharArray();
    chars[index] = replace;
    return String.valueOf(chars);       
}
public static HashSet<Integer> primeFactors(int n) 
{   
    HashSet<Integer> h = new HashSet<Integer>();
    // Print the number of 2s that divide n 
    while (n%2==0) 
    { 
        h.add(2);
        n /= 2; 
    } 


    // n must be odd at this point.  So we can 
    // skip one element (Note i = i +2) 
    for (int i = 3; i <= Math.sqrt(n); i+= 2) 
    { 
      
        // While i divides n, print i and divide n 
        while (n%i == 0) 
        {  
            h.add(i);  
            n /= i; 
        } 
    } 
     
    // This condition is to handle the case whien 
    // n is a prime number greater than 2 
    if(n>2)
      h.add(n);

   return h ;    
     
 }
public static boolean isPrime(int n) 
    { 
        // Corner cases 
        if (n <= 1) 
            return false; 
        if (n <= 3) 
            return true; 
  
        // This is checked so that we can skip 
        // middle five numbers in below loop 
        if (n % 2 == 0 || n % 3 == 0) 
            return false; 
  
        for (int i = 5; i * i <= n; i = i + 6) 
            if (n % i == 0 || n % (i + 2) == 0) 
                return false; 
  
        return true; 
    } 

}