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.util.*;

public class C649_PB {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in = new Scanner(System.in);
		int t = in.nextInt();
		while(t-->0)
		{
			int n = in.nextInt();
			int[] a = new int[n];
			for(int i= 0;i<n;i++)
				a[i] = in.nextInt();
			int dir = 0;
			if(a[1]>a[0])
				dir=1;
			else
				dir=-1;
			ArrayList<Integer> ans = new ArrayList<Integer>();
			ans.add(a[0]);
			for(int i = 1;i<n-1;i++)
			{
				//System.out.println(i+": "+a[i]+" "+a[i+1]);
				if(a[i+1]>a[i])
				{
					if(dir==-1)
					{
						ans.add(a[i]);
						dir =1;
					}
				}
				else if(a[i+1]<a[i])
				{
					if(dir==1)
					{
						ans.add(a[i]);
						dir=-1;
					}
				}
			}
			ans.add(a[n-1]);
			System.out.println(ans.size());
			for(int i =0;i<ans.size()-1;i++)
			{
				System.out.print(ans.get(i)+" ");
			}
			System.out.println(ans.get(ans.size()-1));
		}
		
	}

}