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.*;
import java.util.*;

public class CfA {

    static class Reader {
        BufferedReader br;
        StringTokenizer st;

        public Reader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    static long binpow(int a, int b) {
        if (b == 0)
            return 1;
        long res = binpow(a, b / 2);
        if (b % 2 == 1)
            return res * res * a;
        else
            return res * res;
    }

    static long gcd(long a, long b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    static int n, m, x, y, k;
    static int[] p;
    static int[][] dp;

    public static void main(String[] args) throws Exception {
        Reader in = new Reader();
        int T = in.nextInt();

        while(T-- > 0)
        {   
            n = in.nextInt();
           

            p = new int[n];
            dp = new int[n][2];
            
            dp[0][0] = 0;
            dp[0][1] = -1;
            p[0] = in.nextInt();
            p[1] = in.nextInt();

            int last_ele = 0;
            boolean inc = false;
            List<Integer> list = new ArrayList<>();
            Stack<Integer> stack = new Stack<>();
            stack.push(p[0]);
            stack.push(p[1]);

            if(p[0] < p[1])
                inc = true;
            else
                inc = false;
                    
            for(int i = 2;i < n;i++)
            {
                p[i] = in.nextInt();
                if(p[i] > stack.peek()){
                    if(inc == false)
                        stack.push(p[i]);
                    else{    
                        stack.pop();
                        stack.push(p[i]);
                    }
                    inc = true;
                }
                else{
                    if(inc == true)
                    stack.push(p[i]);
                    else{    
                        stack.pop();
                        stack.push(p[i]);
                    }   
                    inc = false;
                } 
            }

            System.out.println(stack.size());
            for(int s : stack)
                System.out.print(s+" ");
                System.out.println();
            

        }       

    }


}

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