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:
//package com.company;

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
	// write your code here
        Scanner scanner = new Scanner(System.in);
        int cases = scanner.nextInt();

        for(int i = 0; i < cases; i++)
        {
            int l = scanner.nextInt();

            int[] values = new int[l];

            for (int j = 0; j < l; j++ )
            {
                values[j] = scanner.nextInt();
            }

            List<Integer> ans = solve(values);

            System.out.println(ans.size());
            for (Integer a : ans)
                System.out.print(a+" ");

            System.out.println();

        }
    }

    public static List<Integer> solve(int[] values)
    {
        List<Integer> ans = new ArrayList<>();

        ans.add(values[0]);

        for (int i = 1; i < values.length; i++)
        {
            if(values[i] >= values[i-1])
            {
                while (i<values.length && values[i]>=values[i-1])
                    i++;
                ans.add(values[i-1]);
            }

            if(i < values.length && values[i] < values[i-1])
            {
                while (i<values.length && values[i]<=values[i-1])
                    i++;
                ans.add(values[i-1]);
            }
            i--;
        }

        return ans;
    }
}