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.util.*;
import java.math.*;
public class Solution {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int t = in.nextInt();
		for(int tt = 0; tt < t; tt++) {
			int n = in.nextInt();
			int[] ar = new int[n];
			boolean[] flag = new boolean[n];
			for(int i = 0; i < n; i++) {
				ar[i] = in.nextInt();
				flag[i] = true;
			}
			int ans = n;
			int l = 0;
			for(int i = 1; i < n - 1; i++) {
				int r = i + 1;
				if((ar[i] > ar[l]) && (ar[i] > ar[r])) {
					l = i;
				}else if((ar[i] < ar[l]) && (ar[i] < ar[r])) {
					l = i;
				}else {
					flag[i] = false;
					ans -= 1;
				}
			}
			System.out.println(ans);
			for(int i = 0; i < n; i++) {
				if((i > 0) && (flag[i] == true)) System.out.print(" ");
				if(flag[i] == true) System.out.print(ar[i]);
			}
			System.out.print("\n");
		}
	}
}