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.*;
import java.io.*;

public class B {
    public static int compare(int a, int b) {
        if (a>b) 
            return -1;
        else
            return 1;
    }
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(br.readLine());
        while (t>0) {
            int n  =Integer.parseInt(br.readLine());
            ArrayList<Integer> arr = new ArrayList<Integer>();
            ArrayList<Integer> ans = new ArrayList<Integer>();
            String[] line = br.readLine().split(" ");
            for (String s: line) 
                arr.add(Integer.parseInt(s));
            int par = 0;
            if (arr.get(0) > arr.get(1))
                par = -1;
            else    
                par = 1;
            ans.add(arr.get(0));
            for (int i=1 ;i<arr.size(); i++) {
                int c = compare(arr.get(i-1), arr.get(i));
                if (c!=par) {
                    ans.add(arr.get(i-1));
                    par = c;
                }
            }
            ans.add(arr.get(arr.size() - 1));
            System.out.print(ans.size() + "\n");
            for (int i=0; i<ans.size(); i++)
                System.out.print(ans.get(i) + " ");
            System.out.print("\n");
            t--;
        }
    }
}