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.*;
import java.io.*;


public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		
		for(int i=1;i<=t;i++) {
			int n = sc.nextInt();
			int []ar = new int[n+1];
			
			for(int j=1;j<=n;j++) {
				ar[j] = sc.nextInt();
			}
			
			//Vector v = new Vector();
			
			ArrayList<Integer> v = new ArrayList<Integer>();
			
			v.add(ar[1]);
			
			for(int k=2;k<n;k++) {
				if(ar[k]<ar[k-1] && ar[k]<ar[k+1])v.add(ar[k]);
				if(ar[k]>ar[k-1] && ar[k]>ar[k+1])v.add(ar[k]);
			}
			v.add(ar[n]);
			
			System.out.println(v.size());
			
			//for(int x=0;x<v.size();x++)System.out.print(v[x]+" ");
			//System.out.print(v+" ");
			
			for(int x : v) {
				System.out.print(x+" ");
			}
			System.out.println();
		}
	}
}