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.util.*;
import java.util.Map.Entry;
import java.util.concurrent.atomic.AtomicBoolean;
import java.util.stream.Collectors;

public class Main {

  public static void main(String[] args) {
    MyScanner sc = new MyScanner();
    out = new PrintWriter(new BufferedOutputStream(System.out));

    // Start writing your solution here. -------------------------------------

    int t = sc.nextInt();
    while(t-- > 0) {
      int n = sc.nextInt();
      int[] a = new int[n];
      List<Integer> ret = new ArrayList<>();
      int next = -1;
      boolean isIncrease = false;
      for (int i = 0; i < n; i++) {
        a[i] = sc.nextInt();
        if (i == 0) {
          ret.add(a[i]);
          next = a[i];
        } else if (i == 1) {
            isIncrease = a[i] > a[0];
            next = a[i];
        } else {
          if (isIncrease) {
            if (a[i] <= next) {
              ret.add(next);
              isIncrease = false;
            }
          } else {
            if (a[i] >= next) {
              ret.add(next);
              isIncrease = true;
            }
          }
          next = a[i];
        }
      }
      ret.add(next);
      out.println(ret.size());
      for (int i = 0; i < ret.size(); i++) {
        out.print(ret.get(i));
        out.print(" ");
      }
      out.println();
    }

    // Stop writing your solution here. -------------------------------------
    out.close();
  }


  //-----------PrintWriter for faster output---------------------------------
  public static PrintWriter out;

  //-----------MyScanner class for faster input----------
  public static class MyScanner {

    BufferedReader br;
    StringTokenizer st;

    public MyScanner() {
      br = new BufferedReader(new InputStreamReader(System.in));
    }

    String next() {
      while (st == null || !st.hasMoreElements()) {
        try {
          st = new StringTokenizer(br.readLine());
        } catch (IOException e) {
          e.printStackTrace();
        }
      }
      return st.nextToken();
    }

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

    long nextLong() {
      return Long.parseLong(next());
    }

    double nextDouble() {
      return Double.parseDouble(next());
    }

    String nextLine() {
      String str = "";
      try {
        str = br.readLine();
      } catch (IOException e) {
        e.printStackTrace();
      }
      return str;
    }

  }
  //--------------------------------------------------------
}