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.math.BigInteger;
import java.util.*;
 
public class B {
	
	public static void main(String[] args) throws IOException{
		FastScanner sc = new FastScanner();
		PrintWriter pw = new PrintWriter(System.out);
		int yo = sc.nextInt();
		while(yo-->0) {
			int n = sc.nextInt();
			int[] arr = sc.readArray(n);
			
			int index[] = new int[n];
			int count = n;
			for(int i = 1; i < n-1; i++) {
				if((arr[i] < arr[i+1] && arr[i] < arr[i-1]) || 
						(arr[i] > arr[i+1] && arr[i] > arr[i-1])) {
					
				}
				else {
//					arr[i] = arr[i-1];
					index[i] = -1;
					count--;
				}
			}
			pw.println(count);
			for(int i = 0; i < n; i++) {
				if(index[i] != -1) pw.print(arr[i] + " ");
			}
			pw.println();
			
		}
		pw.close();
	}
 
	
 
	static class Pair{
		int x;
		int y;
		public Pair(int x, int y) {
			this.x = x;
			this.y = y;
		}
	}
	
	static int mod = 1000000007;
	
	static long pow(int a, int b) {
		if(b == 0) {
			return 1;
		}
		if(b == 1) {
			return a;
		}
		if(b%2 == 0) {
			long ans = pow(a,b/2);
			return ans*ans;
		}
		else {
			long ans = pow(a,(b-1)/2);
			return a * ans * ans;
		}
		
	}
	
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
	
	static int gcd(int a, int b) {
		return a%b == 0 ? b : gcd(b,a%b);
	}
	
	static boolean[] sieve(int n) {
		
		boolean isPrime[] = new boolean[n+1];
		for(int i = 2; i <= n; i++) {
			if(isPrime[i]) continue;
			for(int j = 2*i; j <= n; j+=i) {
				isPrime[j] = true;
			}
		
		}
		return isPrime;
		
	}
	
//	For Input.txt and Output.txt
	
//	FileInputStream in = new FileInputStream("input.txt");
//	FileOutputStream out = new FileOutputStream("output.txt");
//	PrintWriter pw = new PrintWriter(out);
//	Scanner sc = new Scanner(in);
	
// 	sc.nextLine() 
	
 
}