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.util.*;
public class d2{
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		StringBuilder str = new StringBuilder();
		while(t--!=0){
			int n = sc.nextInt();
         int a[] = new int[n];
         for(int i=0;i<n;i++){
             a[i] = sc.nextInt();
         }
         ArrayList<Integer> l = new ArrayList<>();
         boolean vis[] = new boolean[n];
        for(int i=1;i<n;i++){
            if(a[i]==a[i-1])continue;
            if(a[i]>a[i-1]){
               if(!vis[i-1]){
               l.add(a[i-1]);
               vis[i-1] = true;
            }
               int j = i+1;
               while(j<n && a[j]>=a[j-1])j++;
               if(!vis[j-1]){
               l.add(a[j-1]);
               vis[j-1] = true;
            }
               i = j-1;
            }
            else if(a[i]<a[i-1]){
               if(!vis[i-1]){
               l.add(a[i-1]);
               vis[i-1] = true;
            }
               int j = i+1;
               while(j<n && a[j]<=a[j-1])j++;
               if(!vis[j-1]){
               l.add(a[j-1]);
               vis[j-1] = true;
            }
               i = j-1;
            }
        }
        str.append(""+l.size()+"\n");
        for(int i=0;i<l.size();i++)str.append(""+l.get(i)+" ");
         str.append("\n");
		}
		System.out.println(str.toString());
	}
}