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.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class A {
    static ArrayList<Integer> [] adj;
    static ArrayList<Integer> temp;
    static int mod = (int) 1e9+7;
    static boolean[] vis = new boolean[(int)1e5];

    public static void main(String[] args) throws IOException {
        FastScanner s = new FastScanner();
        int t = s.nextInt();
//        int t = 1;
        while (t-->0) {
            int n = s.nextInt();
            int[] a = new int[n+1];
            List<Integer> l = new ArrayList<>();
            for(int i=1;i<=n;i++){
                a[i] = s.nextInt();
            }
            for(int i=1;i<=n;i++){
                if(i==1 || i==n || (a[i-1]<a[i])!=(a[i]<a[i+1])){
                    l.add(a[i]);
                }
            }
            System.out.println(l.size());
            for(int i:l){
                System.out.print(i+" ");
            }
            System.out.println();
        }
    }
    //Dont't stick on same approach
    //Check long

    static int lower_bound(int[] arr, int x) {
        int low_limit = 0, high_limit = arr.length, mid = -1;
        while (low_limit < high_limit) {
            mid = (low_limit + high_limit) / 2;
            if (arr[mid] >= x) {
                high_limit = mid;
            } else  {
                low_limit = mid + 1;
            }
        }
        return low_limit + 1;
    }


    static void dfs(int i){
        vis[i]=true;
        temp.add(i);
        List<Integer> Y = adj[i];
        for(int x:Y){
            if(!vis[x])dfs(x);
        }
    }

    static long gcd(long a, long b) throws IOException {
        return (b == 0) ? a : gcd(b, a % b);
    }

    static int gcd(int a, int b) throws IOException {
        return (b == 0) ? a : gcd(b, a % b);
    }

    static void sortr(int[] a) {
        ArrayList<Integer> l = new ArrayList<>();
        for (int i : a)
            l.add(i);
        Collections.sort(l, Collections.reverseOrder());
        for (int i = 0; i < a.length; i++)
            a[i] = l.get(i);
    }

    static void sort(int[] a) {
        ArrayList<Integer> l = new ArrayList<>();
        for (int i : a)
            l.add(i);
        Collections.sort(l);
        for (int i = 0; i < a.length; i++)
            a[i] = l.get(i);
    }

    static void sort(long[] a) {
        ArrayList<Long> l = new ArrayList<>();
        for (long i : a)
            l.add(i);
        Collections.sort(l);
        for (int i = 0; i < a.length; i++)
            a[i] = l.get(i);
    }

    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        String next() {
            while (!st.hasMoreTokens())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }
        long nextLong() {
            return Long.parseLong(next());
        }
    }
}