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 Main
{
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int T = sc.nextInt();
		while (T-- > 0) {
		    int n = sc.nextInt();
		    int[] arr = new int[n];
		    
		    for (int i=0; i<n; i++) {
		        arr[i] = sc.nextInt();
		    }   
		    
		    ArrayList<Integer> res = new ArrayList<>();
		    
		    for (int i=0; i<=n-1; i++)
    		{
                if ((i == 0) || (i == n-1) || arr[i] == (Math.max(Math.max(arr[i-1], arr[i+1]), arr[i])) || arr[i] == (Math.min(Math.min(arr[i-1], arr[i+1]), arr[i]))) {
    			    res.add(arr[i]);
    			}
    		}
    		
    		System.out.println(res.size());
    		for (int i=0; i<res.size(); i++) {
    		    System.out.print(res.get(i) + " ");
    		}
    		System.out.println();
		}
	}
}