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.*;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt(), n, p[];
        for (int i = 0; i < t; i++) {
            n = in.nextInt();
            p = new int[n];
            
            for (int j = 0; j < n; j++){
                p[j] = in.nextInt();
            }
            ArrayList<Integer> a = new ArrayList<Integer>();
            
            for (int j = 0; j < n - 1;) {
                a.add(p[j]);
                
                if (p[ j + 1] > p[j]) {
                    while(j + 1 != n && p[j + 1] > p[j])
                        j++;
                }
                
                else if(p[j + 1] <= p[j]) {
                    while(j + 1 != n && p[j + 1] <= p[j]){
                        j++;
                    }
                }
            }
            
            if (p[n - 1] != a.get(a.size() - 1)){
                a.add(p[n - 1]);
            }
            
            System.out.println(a.size());
            
            for (int j = 0; j < a.size() - 1; j++){
                System.out.print(a.get(j) + " ");
            }
            
            System.out.println(a.get(a.size() - 1));
        }
    }
}