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].
import java.io.*;
import java.util.*;
 

public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Unstoppable solver = new Unstoppable();
	  int t=in.nextInt();
	  while(t-->0)
        solver.solve(in, out);
        out.close();
    }
 
    static class Unstoppable { 

       public void solve(InputReader in, PrintWriter out) {
       int n=in.nextInt();
       int a[]=new int[n];
       for(int i=0;i<n;i++){ a[i]=in.nextInt(); }
       boolean b[]=new boolean[n]; b[0]=true;
       int p=0; if(a[1]<a[0]) p=1;
       int count=2;
       for(int i=1;i<n;i++){
           if(a[i]>a[i-1]&&p==0) continue;
            if(a[i]<a[i-1]&&p==1) continue;     
             b[i-1]=true; p=1-p; count++;
       } b[n-1]=true;
       out.println(count);
       for(int i=0;i<n;i++){
               if(b[i]==true) out.print(a[i]+" ");
           }out.println("");
       }
       
       
       
    }
    
 

 
    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
 
        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }
 
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        public long nextLong(){
            return Long.parseLong(next());
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
 
    }
}
   