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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.LinkedList;

public class MostSociallyDistancedSubsequence {

    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        final int t = Integer.parseInt(reader.readLine());


        for(int test = 0; test < t; test++) {
            int n = Integer.parseInt(reader.readLine());
            int[] arr = new int[n];

            String[] inputString = reader.readLine().split(" ");
            for(int i = 0; i < inputString.length; ++i) {
                arr[i] = Integer.parseInt(inputString[i]);
            }
            LinkedList<Integer> res = new LinkedList<>();
            res.add(arr[0]);
            res.add(arr[1]);
            int p = 0;
            int l = 1;
            int current = 2;
            while(current < arr.length) {
                if(Math.abs(arr[l] - arr[p]) + Math.abs(arr[current] - arr[l]) <= Math.abs(arr[p] - arr[current])) {
                    res.pollLast();
                    res.add(arr[current]);
                    l = current;
                } else {
                    p = l;
                    l = current;
                    res.add(arr[current]);
                }
                current++;
            }

            System.out.println(res.size());
            for(Integer v : res) {
                System.out.print(v + " ");
            }
            System.out.println();
        }
    }
}