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.io.*;

public class aaaaaaaa {
	public void run() throws Exception {
		FastReader file = new FastReader();
		int times = file.nextInt();
		for (int asdf = 0; asdf < times; asdf++) {
			int n = file.nextInt();
			ArrayList<Integer> x = new ArrayList();
			for (int i= 0; i < n; i++) x.add(file.nextInt());
			ArrayList<Integer> t = new ArrayList();
			int start = 0;
			int c = 0;
			int cur = x.get(0);
			for (int i = 1; i < n; i++) {
				 if (c == 0) {
					if (x.get(i) > x.get(i - 1)) {
						start = 0;
					}
					else {
						start = 1;
					}
					c = 1;
				}
				else {
					if (x.get(i) > x.get(i - 1) && start != 0) {
						if (cur == x.get(0)) t.add(cur);
						t.add(x.get(i - 1));
						cur = x.get(i);
						start = 0;
					}
					else if (x.get(i) < x.get(i - 1) && start != 1) {
						if (cur == x.get(0)) t.add(cur);
						t.add(x.get(i - 1));
						cur = x.get(i-1);
						start = 1;
					}
				}
			}
			//System.out.println(cur);
			if (cur == x.get(0)) t.add(cur);
			t.add(x.get(n - 1));
			System.out.println(t.size());
			for (int i : t) System.out.print(i + " ");
			System.out.println();
		}
	}

	public static void main(String[] args) throws Exception {
		new aaaaaaaa().run();
	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
}