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.io.*;
import java.math.*;
import java.awt.geom.*;

import static java.lang.Math.*;

public class Solution implements Runnable {

	long mod1 = (long) 1e9 + 7;
	int mod2 = 998244353;

	public void solve() throws Exception {
		int n=sc.nextInt();
		long arr[]=new long[n];
		HashSet<Long> set=new HashSet<Long>();
		for(int i=0; i<n; i++) {
			arr[i]=sc.nextLong();
			set.add(arr[i]);
		}
		long powers[]=new long[65];
		powers[32]=1;
		for(int i=33;i<=64;i++) {
			powers[i]=powers[i-1]*2;
		}
		for(int i=0;i<=31;i++) {
			powers[i]=-1l*powers[64-i];
		}
//		out.println(Arrays.toString(powers));
		ArrayList<Long> three=new ArrayList<Long>();
		ArrayList<Long> two=new ArrayList<Long>();
		ArrayList<Long> one=new ArrayList<Long>();
		one.add(arr[0]);
		for(int i=0;i< n;i++) {
			for(int j=0;j<=63;j++) {
				if(arr[i]+powers[j]>(long)(1e9+2)) break;
				if(three.size()==0 && set.contains(arr[i]+powers[j]) && set.contains(arr[i]+powers[j]*2l)) {
//					if(arr[i]==786007843) {
//						out.println(arr[i]+" "+powers[j]+" "+powers[j+1]);
//					}
					out.println(3);
					out.println(arr[i]+" "+(arr[i]+powers[j])+" "+(arr[i]+powers[j]*2));
					return;
				}
				if(two.size()==0 && set.contains(arr[i]+powers[j])) {
					two.add(arr[i]);
					two.add(arr[i]+powers[j]);
				}
			}
		}
		if(two.size()!=0) {
			out.println(2);
			out.println(two.get(0)+" "+two.get(1));
			return;
		}
		out.println(1);
		out.println(one.get(0));
		

	}

	static long gcd(long a, long b) {
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}

	static void sort(long[] a) {
		ArrayList<Long> l = new ArrayList<>();
		for (long i : a)
			l.add(i);
		Collections.sort(l);
		for (int i = 0; i < a.length; i++)
			a[i] = l.get(i);
	}

	static long ncr(int n, int r, long p) {
		if (r > n)
			return 0l;
		if (r > n - r)
			r = n - r;

		long C[] = new long[r + 1];

		C[0] = 1;

		for (int i = 1; i <= n; i++) {

			for (int j = Math.min(i, r); j > 0; j--)
				C[j] = (C[j] + C[j - 1]) % p;
		}
		return C[r] % p;
	}

	void sieveOfEratosthenes(boolean prime[], int size) {
		for (int i = 0; i < size; i++)
			prime[i] = true;

		for (int p = 2; p * p < size; p++) {
			if (prime[p] == true) {
				for (int i = p * p; i < size; i += p)
					prime[i] = false;
			}
		}
	}

	static int LowerBound(int a[], int x) { // smallest index having value >= x; returns 0-based index
		int l = -1, r = a.length;
		while (l + 1 < r) {
			int m = (l + r) >>> 1;
			if (a[m] >= x)
				r = m;
			else
				l = m;
		}
		return r;
	}

	static int UpperBound(int a[], int x) {// biggest index having value <= x; returns 1-based index
		int l = -1, r = a.length;
		while (l + 1 < r) {
			int m = (l + r) >>> 1;
			if (a[m] <= x)
				l = m;
			else
				r = m;
		}
		return l + 1;
	}

	public long power(long x, long y, long p) {
		long res = 1;
		// out.println(x+" "+y);
		x = x % p;
		if (x == 0)
			return 0;

		while (y > 0) {
			if ((y & 1) == 1)
				res = (res * x) % p;
			y = y >> 1;
			x = (x * x) % p;
		}
		return res;
	}

	static Throwable uncaught;

	BufferedReader in;
	FastScanner sc;
	PrintWriter out;

	@Override
	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			sc = new FastScanner(in);
			solve();
		} catch (Throwable uncaught) {
			Solution.uncaught = uncaught;
		} finally {
			out.close();
		}
	}

	public static void main(String[] args) throws Throwable {
		Thread thread = new Thread(null, new Solution(), "", (1 << 26));
		thread.start();
		thread.join();
		if (Solution.uncaught != null) {
			throw Solution.uncaught;
		}
	}

}

class FastScanner {

	BufferedReader in;
	StringTokenizer st;

	public FastScanner(BufferedReader in) {
		this.in = in;
	}

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

	public int nextInt() throws Exception {
		return Integer.parseInt(nextToken());
	}

	public int[] readArray(int n) throws Exception {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nextInt();
		return a;
	}

	public long nextLong() throws Exception {
		return Long.parseLong(nextToken());
	}

	public double nextDouble() throws Exception {
		return Double.parseDouble(nextToken());
	}

}