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.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

/**
 *
 * @author abacaba
 */
public class b {
    public static void main(String[] args) {
        FS in = new FS(System.in);
        PrintWriter out = new PrintWriter(System.out);
        
        int t = in.nextInt(); int c = 0; while(c++<t) {
//            out.print("Case #" + c + ": ");
            new b().solve(in, out);
        }
        
        out.close();
    }
    
    void solve(FS in, PrintWriter out) {
        int n = in.nextInt();
        int[] arr = new int[n];
        for(int i = 0; i < n; i++) {
            arr[i] = in.nextInt();
        }
        
        StringBuilder res = new StringBuilder();
        
        int tot = 2;
        res.append(arr[0] + " ");
        for(int i = 1; i < n-1; i++) if(!(arr[i-1] <= arr[i] && arr[i] <= arr[i+1]) && !(arr[i+1] <= arr[i] && arr[i] <= arr[i-1])) {
            tot++;
            res.append(arr[i] + " ");
        }
        res.append(arr[n-1]);
        
        out.println(tot + "\n" + res);
    }
    
    static class FS {
        BufferedReader in; StringTokenizer tk;
        public FS(InputStream st) {
            in = new BufferedReader(new InputStreamReader(st));
            tk = new StringTokenizer("");
        }
        
        public String next() {
            if(tk.hasMoreTokens()) return tk.nextToken();
            try { tk = new StringTokenizer(in.readLine()); } catch(Exception e) {}
            return next();
        }
        
        public int nextInt() { return Integer.parseInt(next()); }
        public long nextLong() { return Long.parseLong(next()); }
        public double nextDouble() { return Double.parseDouble(next()); }
    }
}