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 cf1364_Div2_B {
	public static void main(String args[]) {
		Scanner in = new Scanner(System.in);
		int tc = in.nextInt();
		for ( ; tc > 0; tc--) {
			int n = in.nextInt();
			int[] arr = new int[n];
			long sum = 0;
			for (int i = 0; i < n; i++) {
				arr[i] = in.nextInt();
			}
			int[] rem = new int[n + 1];
			int i = 0;
			int j = 1;
			int size = n;
			
			while (i < n && j + 1 < n) {
				if (Math.abs(arr[j + 1] - arr[i]) >= Math.abs(arr[j] - arr[i]) + Math.abs(arr[j + 1] - arr[j])) {
					rem[arr[j]] = 1;
					size--;
				}
				else {
					i = j;
				}
				j++;
			}
			
			System.out.println(size);
			for (int k = 0; k < arr.length; k++) {
				System.out.print(rem[arr[k]] == 0 ? arr[k] + " " : "");
			}
			System.out.println();
			
		}
	}
}
