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.util.*;

public class Codechef {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        while(t-->0){
            int n = in.nextInt();
            int[] p = new int[n];
            boolean b[] = new boolean[n];
            ArrayList<Integer> arr = new ArrayList<Integer>();
            for(int i=0;i<n;i++){
                p[i] = in.nextInt();
                if(i>0){
                    if(p[i-1]-p[i]<0){
                        b[i] = true;
                    }
                    else{
                        b[i] = false;
                    }
                }
            }
            int i=0;
            int u=1;
            // System.out.println(23344);
            arr.add(p[0]);
            for(int j=1;j<n;j++){
                if(b[j]){
                    for(i=u;i<n;i++){
                        if(i==n-1){
                            u=n;
                        }
                        if(b[i])    continue;
                        else{
                            arr.add(p[i-1]);
                            u=i;
                            break;   
                        }
                        
                    }
                    
                }
                else{
                    for(i=u;i<n;i++){
                        if(i==n-1){
                            u=n;
                        }
                        if(!b[i])    continue;
                        else{
                            arr.add(p[i-1]);
                            u=i;
                            break;   
                        }
                        
                    }
                }
                j=u-1;
            }
            arr.add(p[n-1]);
            for(i=1;i<arr.size();i++){
                if(arr.get(i)-arr.get(i-1)==0){
                    arr.remove(arr.get(i));
                }
            }
            System.out.println(arr.size());
            for(i=0;i<arr.size();i++){
                System.out.print(arr.get(i)+" ");
            }
            System.out.println();
        }
        in.close();
    }
}