Problem: There are n distinct points on a coordinate line, the coordinate of i-th point equals to x_i. Choose a subset of the given set of points such that the distance between each pair of points in a subset is an integral power of two. It is necessary to consider each pair of points, not only adjacent. Note that any subset containing one element satisfies the condition above. Among all these subsets, choose a subset with maximum possible size.

In other words, you have to choose the maximum possible number of points x_{i_1}, x_{i_2}, ..., x_{i_m} such that for each pair x_{i_j}, x_{i_k} it is true that |x_{i_j} - x_{i_k}| = 2^d where d is some non-negative integer number (not necessarily the same for each pair of points).

Input

The first line contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of points.

The second line contains n pairwise distinct integers x_1, x_2, ..., x_n (-10^9 ≤ x_i ≤ 10^9) — the coordinates of points.

Output

In the first line print m — the maximum possible number of points in a subset that satisfies the conditions described above.

In the second line print m integers — the coordinates of points in the subset you have chosen.

If there are multiple answers, print any of them.

Examples

Input

6
3 5 4 7 10 12


Output

3
7 3 5

Input

5
-1 2 5 8 11


Output

1
8

Note

In the first example the answer is [7, 3, 5]. Note, that |7-3|=4=2^2, |7-5|=2=2^1 and |3-5|=2=2^1. You can't find a subset having more points satisfying the required property.

Solution (in JAVA):

Solution:
import java.util.*;
import java.math.*;
import java.io.*;

public class CF988D { // can only be a max of 3 numbers that satisfy condition?
	public CF988D() {
		FS scan = new FS();
		int n = scan.nextInt();
		long[] a = new long[n];
		Set<Long> set = new HashSet<>();
		for(int i = 0 ; i < n ; i++) {
			a[i] = scan.nextLong();
			set.add(a[i]);
		}
		List<Long> res = new ArrayList<>();
		res.add(a[0]);
		for(int i = 0 ; i < n ; i++) {
			for(int k = 0 ; k < 40 ; k++) {
				long x0 = a[i];
				long x1 = a[i] - (1L << k);
				long x2 = a[i] + (1L << k);
				if(set.contains(x1) && set.contains(x2) && res.size() < 3) {
					res = new ArrayList<>();
					res.add(x0);
					res.add(x1);
					res.add(x2);
				}
				else if(set.contains(x1) && res.size() < 2) {
					res = new ArrayList<>();
					res.add(x0);
					res.add(x1);
				}
				else if(set.contains(x2) && res.size() < 2) {
					res = new ArrayList<>();
					res.add(x0);
					res.add(x2);
				}
			}
		}
		System.out.println(res.size());
		for(int i = 0 ; i < res.size() ; i++)
			System.out.print(res.get(i) + " ");
	}
	class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		public String next() {
			while(!st.hasMoreTokens()) {
				try { st = new StringTokenizer(br.readLine()); }
				catch(Exception e) { e.printStackTrace(); }
			}
			return st.nextToken();
		}
		public int nextInt() { return Integer.parseInt(next()); }
		public long nextLong() { return Long.parseLong(next()); }
	}
	public static void main(String[] args) { new CF988D(); }
}