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.util.*;

public class A {
    static Scanner s = new Scanner(System.in);

    class Pair {
        int val;
        int idx;

        public Pair(int val, int idx) {
            this.val = val;
            this.idx = idx;
        }
    }

    public static void main(String[] args) {
        // int tc=1;
        int tc = s.nextInt();
        int k = 0;
        while (k < tc) {
            solve2();
            k++;
        }
    }

    public static void solve1() {
        int n = s.nextInt();
        int x = s.nextInt();
        int[] arr = new int[n];
        int sum = 0;
        int d = 0;
        for (int i = 0; i < n; i++) {
            arr[i] = s.nextInt();
            sum += arr[i];
            if (arr[i] % x == 0) {
                d++;
            }
        }

        if (d == n) {
            System.out.println(-1);
        } else if (sum % x != 0) {
            System.out.println(n);
        } else {
            int max = 0;
            int total = sum;
            for (int i = 0; i < n; i++) {
                total -= arr[i];
                if (total % x != 0) {
                    max = Math.max(max, n - i - 1);
                    break;
                }
            }
            total = sum;
            for (int i = n - 1; i >= 0; i--) {
                total -= arr[i];
                if (total % x != 0) {
                    max = Math.max(max, i);
                    break;
                }
            }
            System.out.println(max);

        }
    }
    public static void solve2() {
        int n = s.nextInt();
        int[] arr = new int[n];

        for (int i = 0; i < n; i++) {
            arr[i] = s.nextInt();
            
        }
        ArrayList<Integer> arr1 = new ArrayList<Integer>();

        arr1.add(arr[0]);
        int i=1;
        while(i<n){
            if(arr[i-1]>arr[i]){
                while(i<n&&arr[i-1]>arr[i]){
                    i++;
                }
                arr1.add(arr[i-1]);
            }else{
                while(i<n&&arr[i-1]<arr[i]){
                    i++;
                }
                arr1.add(arr[i-1]);
            }
        }
        System.out.println(arr1.size());
        for(int j=0;j<arr1.size();j++){
            System.out.print(arr1.get(j)+" ");
        }
        System.out.println();
        
    }

}