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.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class DistanceSub {
    public static void main(String args[]) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        int t = Integer.parseInt(br.readLine());

        while (t > 0) {
            t--;

            int n = Integer.parseInt(br.readLine());

            StringTokenizer st = new StringTokenizer(br.readLine());
            int[] a = new int[n];

            for(int i = 0; i < n; i++) a[i] = Integer.parseInt(st.nextToken());

            ArrayList<Integer> sol = new ArrayList<>();
            sol.add(a[0]);

            boolean isIncreasing = (a[1] - a[0] > 0);
            for(int i = 1; i < n; i++){
                int diff = a[i] - a[i-1];

                if((diff > 0 && !isIncreasing) || (diff < 0 && isIncreasing)){
                    sol.add(a[i-1]);
                    isIncreasing = !isIncreasing;
                }
            }
            sol.add(a[n-1]);

            System.out.println(sol.size());
            for(int i  = 0; i < sol.size(); i++){
                System.out.print(sol.get(i) + " ");
            }
        }
    }
}
