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.*;
import java.util.*;

public class Problem1364B {
    public static void main (String args[]) throws IOException {
        BufferedReader input = new BufferedReader(new InputStreamReader(System.in));

        for (int tt = Integer.parseInt(input.readLine()), t = 0; t < tt; t++) { // tt : total testcase
            int n = Integer.parseInt(input.readLine());

            int[] perm = new int[n];
            String[] line = input.readLine().split(" ");
            for (int inx = 0; inx < n; inx++) perm[inx] = Integer.parseInt(line[inx]);

            LinkedList<Integer> answer = new LinkedList<Integer>();
            answer.addLast(perm[0]);

            boolean isAscending = false, wasAscending = false;
            for (int inx = 1; inx < n; inx++) {
                if (perm[inx] < perm[inx - 1]) isAscending = true;
                else isAscending = false;

                if (inx != 1 && isAscending != wasAscending) answer.addLast(perm[inx - 1]);

                wasAscending = isAscending;
            }
            answer.addLast(perm[n - 1]);

            System.out.println(answer.size());
            while (!answer.isEmpty()) {
                System.out.print(answer.pollFirst());
                if (!answer.isEmpty()) System.out.print(" ");
            }
            System.out.println();
        }
    }
}