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.math.*;
import java.util.*;
public class Codeforces{
 
    static int gcd(Integer a,Integer b)
    {
        if(b==0)
            return a;
        else
            return gcd(b,a%b);
    }
 
    static boolean checkForPrime(int inputNumber)
    {
        boolean isItPrime = true;
        if(inputNumber <= 1) 
        {
            isItPrime = false;
            return isItPrime;
        }
        else
        {
            for (int i = 2; i<= inputNumber/2; i++) 
            {
                if ((inputNumber % i) == 0)
                {
                    isItPrime = false;
                    break;
                }
            }
        }
 
        return isItPrime;
    }
 
    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        for(int q=0;q<t;q++)
        {
            int n=sc.nextInt();
            ArrayList<Integer> arr=new ArrayList<Integer>();
            for(int i=0;i<n;i++)
            {
                arr.add(sc.nextInt());
            }
            ArrayList<Integer> ans=new ArrayList<Integer>();
            for(int i=0;i<n;i++)
            {
                if(i==0 || i==n-1 || (arr.get(i)<arr.get(i-1) && arr.get(i)<arr.get(i+1)) || (arr.get(i)>arr.get(i-1) && arr.get(i)>arr.get(i+1)))
                {
                    ans.add(arr.get(i));
                }
            }
            System.out.println(ans.size());
            for(int i=0;i<ans.size();i++)
            {
                System.out.print(ans.get(i)+" ");
            }
            System.out.println();
        }
        sc.close();
    }
}