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.io.PrintWriter;
import java.util.ArrayList;
import java.util.Comparator;
import java.util.List;
import java.util.StringTokenizer;

public class MostSocDistedSubsequence { // Template for CF
    public static class ListComparator implements Comparator<List<Integer>> {

        @Override
        public int compare(List<Integer> l1, List<Integer> l2) {
            for (int i = 0; i < l1.size(); ++i) {
                if (l1.get(i).compareTo(l2.get(i)) != 0) {
                    return l1.get(i).compareTo(l2.get(i));
                }
            }
            return 0;
        }
    }

    public static void main(String[] args) throws IOException {
        // Check for int overflow!!!!
        // Should you use a long to store the sum or smthn?
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);

        int T = Integer.parseInt(f.readLine());
        for (int i = 0; i < T; i++) {
            int N = Integer.parseInt(f.readLine());
            List<Integer> list = new ArrayList<>();
            StringTokenizer st = new StringTokenizer(f.readLine());
            for (int j = 0; j < N; j++) {
                int a = Integer.parseInt(st.nextToken());
                list.add(a);
            }
            List<Integer> best = new ArrayList<>();
            best.add(list.get(0));
            int prev = list.get(0);
            boolean dec = true;
            if (list.get(1) > prev) {
                dec = false;
            }
            for (int j = 1; j < list.size(); j++) {
                if (list.get(j) > prev && dec) {
                    best.add(prev);
                    dec = false;
                } else if (list.get(j) < prev && !dec) {
                    best.add(prev);
                    dec = true;
                }
                prev = list.get(j);
            }
            best.add(list.get(list.size() - 1));
            out.println(best.size());
            for (int j = 0; j < best.size(); j++) {
                out.print(best.get(j) + " ");
            }
            out.println();
        }
        out.close();
    }

}