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.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.StringTokenizer;

public class Solution {
	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while(t-->0) {
			int n =sc.nextInt();
			int arr[]= new int[n];
			for(int i=0;i<n;i++) {
				arr[i]=sc.nextInt();
			}
			ArrayList<Integer> ans = new  ArrayList<Integer>();
			int curr = arr[0];
			boolean greater = arr[1]>arr[0];
			ans.add(curr);
			for(int i=1;i<n-1;i++) {
				if(greater && arr[i]>arr[i+1]) {
					ans.add(arr[i]);
					greater = false;
				}else if(!greater && arr[i]<arr[i+1]) {
					ans.add(arr[i]);
					greater = true;
				}
			}
			ans.add(arr[n-1]);
			System.out.println(ans.size());
			StringBuilder sb = new StringBuilder();
			for(int x: ans) {
				sb.append(x+" ");
			}
			System.out.println(sb.toString().trim());
		}
	}
}

class Scanner {
	StringTokenizer st;
	BufferedReader br;

	public Scanner(InputStream s) {
		br = new BufferedReader(new InputStreamReader(s));
	}

	public String next() throws IOException {
		while (st == null || !st.hasMoreTokens())
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}

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

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

	public String nextLine() throws IOException {
		return br.readLine();
	}

	public double nextDouble() throws IOException {
		String x = next();
		StringBuilder sb = new StringBuilder("0");
		double res = 0, f = 1;
		boolean dec = false, neg = false;
		int start = 0;
		if (x.charAt(0) == '-') {
			neg = true;
			start++;
		}
		for (int i = start; i < x.length(); i++)
			if (x.charAt(i) == '.') {
				res = Long.parseLong(sb.toString());
				sb = new StringBuilder("0");
				dec = true;
			} else {
				sb.append(x.charAt(i));
				if (dec)
					f *= 10;
			}
		res += Long.parseLong(sb.toString()) / f;
		return res * (neg ? -1 : 1);
	}

	public boolean ready() throws IOException {
		return br.ready();
	}

}