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.text.*;

public class Main{
    final long mod = (long)1e9+7, IINF = (long)1e19;
//    final int MAX = (int)1e5+1, MX = (int)1e7+1, INF = (int)1e9;
    DecimalFormat df = new DecimalFormat("0.0000000000");
    final double EPS = 1e-8;
    FastReader in;
    PrintWriter out;
    boolean multipleTC = false;

    public static void main(String[] args) throws Exception{new Main().run();};
    void run() throws Exception{
        in = new FastReader();
        out = new PrintWriter(System.out);
        for(int i = 1, t = (multipleTC)?ni():1; i<=t; i++)solve(i);
        out.flush();    
        out.close();
    }

    void solve(int TC){
        int n = ni();
        long[] a = new long[n];
        HashSet<Long> s = new HashSet<>();
        for(int i = 0; i< n; i++){
            a[i] = nl();
            s.add(a[i]);
        }
        sort(a);
        int ans = 1;
        long[] a2 = new long[2], a3 = new long[3];
        for(int i = 0; i< n; i++){
            long D = 1;
            while(a[n-1]>= a[i]+D && ans<3){
                if(s.contains(a[i]+D) && s.contains(a[i]+(D<<1))){
                    ans = 3;
                    a3[0] = a[i];
                    a3[1] = a[i]+D;
                    a3[2] = a[i]+(D<<1);
                }else if(ans==1 && s.contains(a[i]+D)){
                    ans = 2;
                    a2[0] = a[i];
                    a2[1] = a[i]+D;
                }
                D*=2;
            }
        }
        pn(ans);
        if(ans==1)pn(a[0]);
        else if(ans==2)pn(a2[0]+" " + a2[1]);
        else if(ans==3)pn(a3[0]+" " + a3[1]+" " + a3[2]);
    }
    
    int[] reverse(int[] a){
        int[] o = new int[a.length];
        for(int i = 0; i< a.length; i++)o[i] = a[a.length-i-1];
        return o;   
    }

    int[] sort(int[] a){
        if(a.length==1)return a;
        int mid = a.length/2;
        int[] b = sort(Arrays.copyOfRange(a,0,mid)), c = sort(Arrays.copyOfRange(a,mid,a.length));
        for(int i = 0, j = 0, k = 0; i< a.length; i++){
            if(j<b.length && k<c.length){
                if(b[j]<c[k])a[i] = b[j++];
                else a[i] = c[k++];
            }else if(j<b.length)a[i] = b[j++];
            else a[i] = c[k++];
        }
        return a;
    }

    long[] sort(long[] a){
        if(a.length==1)return a;
        int mid = a.length/2;
        long[] b = sort(Arrays.copyOfRange(a,0,mid)), c = sort(Arrays.copyOfRange(a,mid,a.length));
        for(int i = 0, j = 0, k = 0; i< a.length; i++){
            if(j<b.length && k<c.length){
                if(b[j]<c[k])a[i] = b[j++];
                else a[i] = c[k++];
            }else if(j<b.length)a[i] = b[j++];
            else a[i] = c[k++];
        }
        return a;
    }

    int[] ia(int ind,int n){
        int[] out = new int[ind+n];
        for(int i = 0; i< n; i++)out[ind+i] = ni();
        return out;
    }

    long[] la(int ind, int n){
        long[] out = new long[ind+n];
        for(int i = 0; i< n; i++)out[ind+i] = nl();
        return out;
    }
    
    double[] da(int ind, int n){
        double[] out = new double[ind+n];
        for(int i = 0; i< n; i++)out[ind+i] = nd();
        return out;
    }

    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bitcount(long n){return (n==0)?0:(1+bitcount(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n(){return in.next();}
    String nln(){return in.nextLine();}
    int ni(){return Integer.parseInt(in.next());}
    long nl(){return Long.parseLong(in.next());}
    double nd(){return Double.parseDouble(in.next());}

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

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

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

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