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.util.*;
import java.util.Map.Entry;
import java.lang.*;
import java.math.*;
import java.text.*;
import java.io.*;
		public final class Solve {
		static PrintWriter out = new PrintWriter(System.out); 
		static void flush() {
	 	out.flush();
	 	}
		static void run(long s, long e) {
		NumberFormat formatter = new DecimalFormat("#0.00000");
		System.out.print("Execution time is " + formatter.format((e - s) / 1000d) + " seconds");
		}
		static int gcd(int a, int b) { 
			if (b == 0) return a; 
			return gcd(b, a % b);  
		}
		static long gcd(long a, long b) { 
			if (b == 0) return a; 
			return gcd(b, a % b);  
		}
		static long lcm(long a, long b) { 
			return a*b/gcd(a, b);  
		}
		static class FastReader {
	     static BufferedReader br;
	    StringTokenizer st;
	
	    public FastReader() {
	    br = new BufferedReader(new
	    InputStreamReader(System.in));
	    }
	    static boolean isPalindrome(String str1, String str2) {
	    String str3 = str1+str2;
	    int i = 0, j = str3.length()-1;
	    while(i < j) {
	    char a = str3.charAt(i), b = str3.charAt(j);
	    if(a != b) return false;
	    i++;j--;
	    }	
	    return true;
	    }
	    static boolean isPalindrome(String str) {
	    int i = 0, j = str.length()-1;
	    while(i < j) {
	    char a = str.charAt(i), b = str.charAt(j);
	    if(a != b) return false;
	    i++;j--;
	    }
	    return true;
	    }
	    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());}
	    static int fact(int n) {
	    if(n == 1) return 1;
	    return n * fact(n-1);
	   	}
	    public int[] readIntArray(int n) {
	    int[] arr = new int[n];
		for(int i=0; i<n; ++i)
		arr[i]=nextInt();
		return arr;
		}
	    public long[] readLongArray(int n) {
	    	long[] arr = new long[n];
	    	for(int i=0; i<n; ++i)
	    		arr[i]=nextLong();
	    	return arr;
	    }
	    public int[][] readIntArray(int m, int n){
		int[][] arr = new int[m][n];
		for(int i = 0;i<m;i++) 
		for(int j = 0;j<n;j++) 
		arr[i][j] = nextInt();
		return arr;
	    }
	    public String[] readStringArray(int n) {
		String[] arr = new String[n];
		for(int i=0; i<n; ++i)
		arr[i]= nextLine();
		return arr;	
	    }
	    double nextDouble() {return Double.parseDouble(next());}
	    String nextLine() {
	    String str = "";
	    try{str = br.readLine();}
	    catch (IOException e) {e.printStackTrace();}
	    return str;}
	    }
		static void solve(int[] arr, int n){
			long sum = 0;
			int cind = 0, c = 0;
			for(int i = 1;i<n-1;i++) {
				int x = Math.abs(arr[i] - arr[cind]);
				int y = Math.abs(arr[i] - arr[i+1]);
				int z = Math.abs(arr[cind] - arr[i+1]);
				
				if(z >= x && z >= y) {
					arr[i] = 0;
					c++;
					sum += z;
				}
				else {
					sum += x;
					sum += y;
					
					cind = i;
				}
			}
			if(Math.abs(arr[n-1] - arr[0]) >= sum) {
				out.println(2);
				out.println(arr[0]+" "+arr[n-1]);
				return;
			}
			out.println(n-c);
			for(int i = 0;i<n;i++) {
				if(arr[i] != 0) {
					out.print(arr[i]+" ");
				}
			}
			out.println();
		}
		public static void main(String args[]) throws IOException {
	    FastReader sc = new FastReader();
	    long s1 = System.currentTimeMillis();
	    int t = sc.nextInt();
	    while(t-- > 0 ) {
	    	int n = sc.nextInt();
	    	int[] arr = sc.readIntArray(n);
	    	solve(arr, n);
	    }
	    flush();
	    long e = System.currentTimeMillis();
//	    run(s1,e);
	    }
	}