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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class B_Most_Socially_Distanced_Subsequence {
	public static void main(String[] args) {
		MyScanner sc = new MyScanner();
	    PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
	    
	    int t = sc.nextInt();
	    
	    while(t-- > 0) {
	    	int n = sc.nextInt();
	    	
	    	int[] vals = new int[n];
	    	for(int i = 0; i < n; i++) {
	    		vals[i] = sc.nextInt();
	    	}
	    	
	    	int subseq_length = n;
	    	// mark with a 1 to indicate removal
	    	boolean[] remove_indexes = new boolean[n];
	    	for(int i = 1; i < n - 1; i++) {
	    		if((vals[i] >= vals[i - 1] && vals[i] <= vals[i + 1])
	    				|| (vals[i] <= vals[i - 1] && vals[i] >= vals[i + 1])) {
	    			subseq_length--;
	    			remove_indexes[i] = true;
	    		}
	    	}
	    	// handle first and last index
	    	if(vals[0] == vals[1]) {
    			subseq_length--;
    			remove_indexes[0] = true;
	    	}
	    	if(vals[n - 1] == vals[n - 2]) {
    			subseq_length--;
    			remove_indexes[n - 1] = true;
	    	}
	    	
	    	out.println(subseq_length);
	    	for(int i = 0; i < remove_indexes.length; i++) {
	    		if(!remove_indexes[i]) {
	    			out.print(vals[i] + " ");
	    		}
	    	}
	    	out.println();
	    }
	    
	    out.close();
	}
	
	private 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;
	   }
	}
}