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.*;
import java.util.*;

public class B { //

    static int[] a = new int[100000];
    static boolean[] in = new boolean[100000];

    public static void main(String[] args) {
        try {
            BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
            BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
            StringBuilder sb = new StringBuilder();
            int testCnt = Integer.parseInt(br.readLine());
            for (int i = 0; i < testCnt; i++) {
                int len = Integer.parseInt(br.readLine());
                String[] line = br.readLine().split(" ");
                for (int j = 0; j < len; j++) {
                    a[j] = Integer.parseInt(line[j]);
                }
                int prev = a[0];
                int cnt = len;
                Arrays.fill(in, true);
                for (int j = 1; j < len - 1; j++) {
                    if ((prev < a[j] && a[j] < a[j + 1]) ||
                            (prev > a[j] && a[j] > a[j + 1])) {
                        cnt--;
                        in[j] = false;
                    }
                    else prev = a[j];
                }
                sb.append(cnt + "\n");
                for (int j = 0; j < len; j++) {
                    if (in[j]) sb.append(a[j] + " ");
                }
                sb.append("\n");
            }
            bw.write(sb.toString());
            br.close();
            bw.close();
        }
        catch (Exception e) {
            e.printStackTrace();
        }
    }

    public static void dbug(String s, long ... a) { // /*
        String[] line = s.split("<>");
        for (int i = 0; i < a.length; i++) {
            System.out.print(line[i] + a[i]);
        }
        System.out.println(line[a.length]); // */
    }

    public static void print(int[] a) { // /*
        for (int i = 0; i < a.length; i++) {
            System.out.print(a[i] + " ");
        }
        System.out.println(); // */
    }
}