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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;



public class my{
public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int t = s.nextInt();
		while (t-- > 0) {
			int n = s.nextInt();
			int[] arr = new int[n];
			for (int i = 0; i < n; i++) {
				arr[i] = s.nextInt();
			}

			incdec(arr, n);
		}
	}

	private static void incdec(int[] arr, int n) {

		int sum = 0;
		boolean dec = true, inc = true;
		StringBuilder str = new StringBuilder();
		Stack<Integer> st = new Stack<>();
		st.push(arr[0]);
		int i = 1;
		str.append(arr[0] + " ");
		while (i < n) {
			if (arr[i] > st.peek()) {
				while (i < n && arr[i] >= st.peek()) {
					sum += Math.abs(st.peek() - arr[i]);
					st.push(arr[i]);
					i++;
				}
				if (i < n)
					str.append(arr[i - 1] + " ");
			} else {
				while (i < n && arr[i] <= st.peek()) {
					sum += st.peek() - arr[i];
					st.push(arr[i]);
					i++;
				}
				if (i < n)
					str.append(arr[i - 1] + " ");
			}
		}
		str.append(arr[n - 1]);
		String[] ste = str.toString().split(" ");
		System.out.println(ste.length);
		System.out.println(str.toString());

	}
}