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.io.*;
import java.util.*;

import static java.lang.Math.max;

public class B {
	Reader source;
	BufferedReader br;
	StringTokenizer in;
	PrintWriter out;
	PrintWriter err;

	void go() throws Exception {
		int n = nextInt();
		int[] p = new int [n];
		p[0] = nextInt();
		int[] d = new int [n];
		for (int i = 1; i < n; i++) {
			p[i] = nextInt();
			d[i] = (p[i] < p[i - 1] ? -1 : +1);
		}
		d[0] = d[1];
		List<Integer> ans = new ArrayList<>();
		ans.add(p[0]);
		for (int i = 2; i < n; i++) {
			if (d[i - 1] != d[i]) {
				ans.add(p[i - 1]);
			}
		}
		ans.add(p[n - 1]);
		out.println(ans.size());
		for (int i : ans) out.print(i + " ");
		out.println();
	}

	void solve() throws Exception {
		for (int T = nextInt(); T > 0; T--) go();
	}
	void run() throws Exception {
		if (OJ) source = new InputStreamReader(System.in);
		else 	source = new FileReader("B.in");
		br = new BufferedReader(source);
		out = new PrintWriter(System.out);
		err = new PrintWriter(System.out);
		solve();
		out.flush();
		err.flush();
	}
	public static void main(String[] args) throws Exception {
		new B().run();
	}
	boolean OJ = System.getProperty("ONLINE_JUDGE") != null;

	void show(int[] a) {
		for (int i = 0; i < a.length; i++)
			err.printf("%4d ", a[i]);
		err.println();
	}
	void show(long[] a) {
		for (int i = 0; i < a.length; i++)
			err.printf("%4d ", a[i]);
		err.printf("\n");
	}

	int lower_bound(int[] a, int l, int r, int x) {
		int m;
		while (l + 1 < r) {
			m = (l + r) / 2;
			if (a[m] < x) l = m;
			else r = m;
		}
		return (a[l] == x) ? l : r;
	}
	int lower_bound(int[] a, int x) {
		return lower_bound(a, 0, a.length, x);
	}
	int lower_bound(long[] a, int l, int r, long x) {
		int m;
		while (l + 1 < r) {
			m = (l + r) / 2;
			if (a[m] < x) l = m;
			else r = m;
		}
		return (a[l] == x) ? l : r;
	}
	int lower_bound(long[] a, long x) {
		return lower_bound(a, 0, a.length, x);
	}
	int upper_bound(int[] a, int l, int r, int x) {
		int m;
		while (l + 1 < r) {
			m = (l + r) / 2;
			if (a[m] <= x) l = m;
			else r = m;
		}
		return (a[l] > x) ? l : r;
	}
	int upper_bound(int[] a, int x) {
		return upper_bound(a, 0, a.length, x);
	}
	int upper_bound(long[] a, int l, int r, long x) {
		int m;
		while (l + 1 < r) {
			m = (l + r) / 2;
			if (a[m] <= x) l = m;
			else r = m;
		}
		return (a[l] > x) ? l : r;
	}
	int upper_bound(long[] a, long x) {
		return upper_bound(a, 0, a.length, x);
	}

	String nextToken() throws Exception {
		while (in == null || !in.hasMoreTokens()) {
			in = new StringTokenizer(br.readLine());
		}
		return in.nextToken();
	}
	int nextInt() throws Exception {
		return Integer.parseInt(nextToken());
	}
	long nextLong() throws Exception {
		return Long.parseLong(nextToken());
	}
	double nextDouble() throws Exception {
		return Double.parseDouble(nextToken());
	}
}
