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 Main
{
	static Scanner scan = new Scanner(System.in);
	public static void solve() throws Exception
	{
		//BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		//StringTokenizer tk = new StringTokenizer(in.readLine());
		int n = scan.nextInt();
		int[] nums = new int[n];
		for(int i=0;i<n;i++)
		{
			nums[i]=scan.nextInt();
		}
		ArrayList<Integer> a = new ArrayList<>();
		a.add(nums[0]);
		for(int i=1;i<n-1;i++)
		{
			if(Math.abs(nums[i+1]-nums[i-1])<Math.abs(nums[i+1]-nums[i])+Math.abs(nums[i]-nums[i-1]))
				a.add(nums[i]);
		}
		a.add(nums[n-1]);
		System.out.println(a.size());
		for(int i=0;i<a.size();i++)
		{
			System.out.print(a.get(i)+" ");
		}
		System.out.println();
	}
	public static void main(String[] args) throws Exception
	{
		if(scan.hasNext())
		{
			int t= scan.nextInt();
			//int t= 1;
			outer:while(t-->0)
			{	
				solve();
			}
		}
	}
}