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.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Scanner;

public class MostSociallyDistancedSubsequence {
    public static void main(String[] args){
        Scanner s = new Scanner(System.in);
        int nTest = s.nextInt();
        for(int t = 0; t< nTest; t++){
            int pSize= s.nextInt();
            ArrayList<Integer> arr = new ArrayList<Integer>();
            for(int i = 0; i<pSize; i++) arr.add(s.nextInt());
            HashSet<Integer> tbRemoved = new HashSet<Integer>();

            for(int i = 0; i<pSize-2; i++){
                int[] temp = new int[3];
                temp[0] = arr.get(i);
                temp[1] = arr.get(i+1);
                temp[2] = arr.get(i+2);
                Arrays.sort(temp);
                if(temp[1] != arr.get(0) && temp[1] != arr.get(pSize-1)){
                    if(temp[1] == arr.get(i+1)){
                        tbRemoved.add(Integer.valueOf(temp[1]));
                    }
                }
            }
            arr.removeAll(tbRemoved);
            System.out.println(arr.size());
            for(int r : arr){
                System.out.print(r+" ");
            }
            System.out.println();
        }
    }
}