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.*;
import java.util.*;

public class B {

	static long m = (long) (1e9 + 7);

	public static void main(String[] args) throws IOException {
		Scanner scn = new Scanner(System.in);
		StringBuilder sb = new StringBuilder();
		int T = scn.nextInt(), tcs = 0;
		B: while (tcs++ < T) {
			int n = scn.nextInt();
			long a[] = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = scn.nextLong();
			int i = 1;
			StringBuilder st1 = new StringBuilder();
			st1.append(a[0] + " ");
			int c = 1;
			while (i < n) {
				if (a[i] > a[i - 1]) {
					while (i < n && a[i] > a[i - 1]) {
						i++;
					}
				} else {
					while (i < n && a[i] < a[i - 1]) {
						i++;
					}
				}
				i--;
				if (i == n)
					break;
				c++;
				st1.append(a[i] + " ");
				i++;

			}
			sb.append(c + "\n");
			sb.append(st1 + "\n");
		}
		System.out.print(sb);
	}

	static class Scanner {

		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

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

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