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 javafx.util.Pair;

import java.util.*;
/*
NOTES:
-more than 10 digits (10^10), use long
-prefix sum of 1's and -1's to control flow of highlighting a region
-log(a)/log(b) = log base b of a
-create a position array if trying to see if elements show up next to each other sequentially
 i.e. elements 1,2,3 are within 3 spots of each other in an array
      arr[scannedInt - 1] = i; (position of i at p[i] in original)
- use a freq array when counting instances of an element occurring
- increment the counter first before simulating anything turn-based (two player game)

 */

public class cf {
    static long mod = (long)(Math.pow(10, 9) + 7);
    public static String alpha = "abcdefghijklmnopqrstuvwxyz";
    public static void main(String[] args) {
        Scanner scan =new Scanner(System.in);
        int t= scan.nextInt();
        while(t-->0){
            int n = scan.nextInt();
            int[] a = new int[n];
            for(int i = 0; i<n;i++){
                a[i] = scan.nextInt();
            }
            ArrayList<Integer> al = new ArrayList<>();
            al.add(a[0]);
            for(int i = 1; i<n-1; i++){
                if(a[i]>a[i-1] && a[i]>a[i+1]) al.add(a[i]);
                if(a[i]<a[i-1] && a[i]<a[i+1]) al.add(a[i]);
            }
            al.add(a[n-1]);
            System.out.println(al.size());
            for(int i : al){
                System.out.print(i+" ");
            }
            System.out.println();
        }
        scan.close();
    }





    static int GCD(int a, int b) {
        // Everything divides 0
        if (a == 0) return b;
        if (b == 0) return a;

        // base case
        if (a == b) return a;

        // a is greater
        if (a > b) return GCD(a-b, b);
        return GCD(a, b-a);
    }
    static int LCM(int a, int b){
        return (a*b)/(GCD(a,b));
    }
    static String rev(String s){
        char[] arr = s.toCharArray();
        for(int i = 0; i<(s.length()+1)/2; i++){
            char temp = arr[i];
            arr[i] = arr[s.length()-i-1];
            arr[s.length()-i-1] = temp;
        }
        String fin = new String(arr);
        return fin;
    }
    static long pow(long a, long N) {
        if (N == 0) return 1;
        else if (N == 1) return a;
        else {
            long R = pow(a,N/2);
            if (N % 2 == 0) {
                return R*R;
            }
            else {
                return R*R*a;
            }
        }
    }

    static long powMod(long a, long N) {
        if (N == 0) return 1;
        else if (N == 1) return a % mod;
        else {
            long R = powMod(a,N/2) % mod;
            R *= R;
            R %= mod;
            if (N % 2 == 1) {
                R *= a;
                R %= mod;
            }
            return R % mod;
        }
    }

}