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].
import java.io.*;
import java.nio.charset.StandardCharsets;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;

public class MostSociallyDistancedSubsequence {
    private static final Console io = new Console();

    public static void main(String[] args) throws IOException {
        for (int t = io.integers(); t > 0; t--) solve();
        io.close();
    }

    private static void solve() throws IOException {
        int n = io.integers();
        int[] list = new int[n];
        for (int i = 0; i < n; i++) list[i] = io.integers();
        int derivative = 0;
        List<Integer> answer = new ArrayList<>();
        for (int i = 1; i < n; i++) {
            int compare = Integer.compare(list[i], list[i - 1]);
            if ((derivative == 0 && compare != 0) ||
                    (derivative != 0 && compare == -1 * derivative))
                answer.add(list[i - 1]);
            if (compare != 0) derivative = compare;
        }
        if (n >= 2 && list[n - 1] != list[n - 2]) answer.add(list[n - 1]);
        io.println(String.valueOf(answer.size()));
        for (int found : answer) io.print(found + " ");
        io.println();
    }

    private static class Console {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in, StandardCharsets.UTF_8));
        StringTokenizer tokenizer = new StringTokenizer("");
        BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out, StandardCharsets.UTF_8));

        String next() {
            while (!tokenizer.hasMoreElements()) {
                try {
                    tokenizer = new StringTokenizer(in.readLine());
                } catch (Exception ignored) {
                }
            }
            return tokenizer.nextToken();
        }

        int integers() {
            return Integer.parseInt(next());
        }

        void println(String s) throws IOException {
            out.write(s + "\n");
        }

        void println() throws IOException {
            out.write("\n");
        }

        void print(String s) throws IOException {
            out.write(s);
        }

        void close() throws IOException {
            in.close();
            out.close();
        }
    }
}
