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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;
public class cdf649b
{
	static void merge(int arr[], int l, int m, int r) 
    { 
        int n1 = m - l + 1; 
        int n2 = r - m; 
        int L[] = new int [n1]; 
        int R[] = new int [n2]; 
        for (int i=0; i<n1; ++i) 
            L[i] = arr[l + i]; 
        for (int j=0; j<n2; ++j) 
            R[j] = arr[m + 1+ j]; 
        int i = 0, j = 0; 
        int k = l; 
        while (i < n1 && j < n2) 
        { 
            if (L[i] <= R[j]) 
            { 
                arr[k] = L[i]; 
                i++; 
            } 
            else
            { 
                arr[k] = R[j]; 
                j++; 
            } 
            k++; 
        } 
        while (i < n1) 
        { 
            arr[k] = L[i]; 
            i++; 
            k++; 
        } 
        while (j < n2) 
        { 
            arr[k] = R[j]; 
            j++; 
            k++; 
        } 
    } 
    static void sort(int arr[], int l, int r) 
    { 
        if (l < r) 
        { 
            int m = (l+r)/2; 
            sort(arr, l, m); 
            sort(arr , m+1, r); 
            merge(arr, l, m, r); 
        } 
    } 
    public static int lowerBound(int[] array, int length, int value) 
	{
        int low = 0;
        int high = length;
        while (low < high) 
        {
            final int mid = (low + high) / 2;
            //checks if the value is less than middle element of the array
            if (value <= array[mid]) 
            {
                high = mid;
            } 
            else 
            {
                low = mid + 1;
            }
        }
        return low;
    }
	public static int upperBound(int[] array, int length, int value) 
	{
        int low = 0;
        int high = length;
        while (low < high) {
            final int mid = (low + high) / 2;
            if (value >= array[mid]) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        return low;
    }
    static int gcd(int a, int b) 
    { 
      if (b == 0) 
        return a; 
      return gcd(b, a % b);  
    } 
    static long power(long n,long m)
    {
    	if(m==0)
    		return 1;
    	long ans=1;
    	while(m>0)
    	{
    		ans=ans*n;
    		m--;
    	}
    	return ans;
    }
    static int BinarySearch(int arr[], int x) 
    { 
        int l = 0, r = arr.length - 1; 
        while (l <= r) { 
            int m = l + (r - l) / 2; 
            if (arr[m] == x) 
                return m; 
            if (arr[m] < x) 
                l = m + 1; 
            else
                r = m - 1; 
        } 
        return -1; 
    } 
	public static void main(String args[])throws Exception
	{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		int t=Integer.parseInt(br.readLine());
	    for(int i=1;i<=t;i++)
	    {
	        int n=Integer.parseInt(br.readLine());
	        int arr[]=new int[n];
	        ArrayList <Integer> arrli=new ArrayList<Integer>();
	        String s=br.readLine();
	        String str[]=s.split(" ");
	        for(int j=0;j<n;j++)
	        arr[j]=Integer.parseInt(str[j]);
	        for(int j=0;j<n;)
	        {
	        	arrli.add(arr[j]);
	        	int k=j+1;
	        	if(k<n&&arr[k]>arr[j])
	        	{
	        	while(k+1<n&&arr[k+1]>arr[k])
	        		k++;
	        	}
	        	else
	        	{
	        		while(k+1<n&&arr[k+1]<arr[k])
		        		k++;
	        	}
	        	j=k;
	        }
	        System.out.println(arrli.size());
	        for(int j=0;j<arrli.size();j++)
	        System.out.print(arrli.get(j)+" ");
	        System.out.println();
	    }
	}
}