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.ArrayList;
import java.util.HashMap;
import java.util.Scanner;

public class sociallyDistanceSub {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int t = s.nextInt();
        for(int i=0; i<t; i++) {
            int n = s.nextInt();
            int arr[] = new int[n];
            int sortedArray[] = new int[n];
            HashMap<Integer, Integer> map = new HashMap<>();
            ArrayList<Integer> good = new ArrayList<>();
            for(int j=0; j < n;j++) {
                int a= s.nextInt();
                arr[j] = a;
                sortedArray[j] = a;
                map.put(a, j);
            }
            good.add(arr[0]);
            for(int z =1; z< n-1; z++) {
                if(arr[z] >= arr[z+1] && arr[z] <= arr[z-1]) {
                    continue;
                }
                else if(arr[z] >= arr[z-1] && arr[z] <= arr[z+1]) {
                    continue;
                }
                else {
                    good.add(arr[z]);
                }
            }
            good.add(arr[n-1]);
            System.out.println(good.size());
            for(int z=0;z < good.size(); z++) {
                System.out.println(good.get(z));
            }






        }
    }
}