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:
//package codeforces._649;

import java.io.*;
import java.util.LinkedList;
import java.util.StringTokenizer;

public class _648_B {

    public static void main (String[] args) throws Exception {

        String s = "2\n" +
                "3\n" +
                "3 2 1\n" +
                "4\n" +
                "1 3 4 2";


//        br = new BufferedReader(new StringReader(s));
        br = new BufferedReader(new InputStreamReader(System.in));
        bw = new BufferedWriter(new OutputStreamWriter(System.out));

        rl(); int T = nin();

        for (int t=0; t<T; ++t) {
            rl(); int n=nin();
            rl(); int A[] = new int[n];
            for(int i=0; i<n; ++i) {
                A[i] = nin();
            }

            LinkedList<Integer> list = new LinkedList<>();
            list.add(A[0]); list.add(A[1]);
            for (int i=2; i<n; ++i) {
                if ( (( ((double)A[i-2])-A[i-1]) / (((double)A[i-1])-A[i]))  > 0 ) {
                    list.removeLast();
                }
                list.add(A[i]);
            }


            bw.write(Integer.toString(list.size()));
            bw.newLine();

            for (int i:list) bw.write(Integer.toString(i)+" ");
            bw.newLine();
        }
        bw.flush();
    }

    static BufferedReader br;
    static BufferedWriter bw;
    static StringTokenizer st;
    static void rl() throws Exception{
        st = new StringTokenizer(br.readLine());
    }
    static long nlo(){
        return Long.parseLong(st.nextToken());
    }
    static int nin(){
        return Integer.parseInt(st.nextToken());
    }
    /*private static void te(){
      }*/
    static double ndo(){
        return Double.parseDouble(st.nextToken());
    }
    static char[] nca(){
        return st.nextToken().toCharArray();
    }
    static String nstr(){
        return st.nextToken();
    }
}