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.*;
public class Main implements Runnable{
    FastScanner sc;
    PrintWriter pw;
    final class FastScanner {
        BufferedReader br;
        StringTokenizer st;
 
        public FastScanner() {
            try {
                br = new BufferedReader(new InputStreamReader(System.in));
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                e.printStackTrace();
            }
        }
 
        public long nlo() {
            return Long.parseLong(next());
        }
 
        public String next() {
            if (st.hasMoreTokens()) return st.nextToken();
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                e.printStackTrace();
            }
            return st.nextToken();
        }
 
        public int ni() {
            return Integer.parseInt(next());
        }
 
        public String nli() {
            String line = "";
            if (st.hasMoreTokens()) line = st.nextToken();
            else try {
                return br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            while (st.hasMoreTokens()) line += " " + st.nextToken();
            return line;
        }
 
        public double nd() {
            return Double.parseDouble(next());
        }
    }
    public static void main(String[] args) throws Exception
    {
        new Thread(null,new Main(),"codeforces",1<<28).start();
    }
    public void run()
    {
        sc=new FastScanner();
        pw=new PrintWriter(System.out);
        solve();
        pw.flush();
        pw.close();
    }
    public long gcd(long a,long b)
    {
        return b==0L?a:gcd(b,a%b);
    }
    public long ppow(long a,long b,long mod)
    {
        if(b==0L)
        return 1L;
        long tmp=1;
        while(b>1L)
        {
            if((b&1L)==1)
            tmp*=a;
            a*=a;
            a%=mod;
            tmp%=mod;
            b>>=1;
        }
        return (tmp*a)%mod;
    }
    public  int gcd(int x,int y)
    {
        return y==0?x:gcd(y,x%y);
    }
   
    //////////////////////////////////
    /////////////  LOGIC  ///////////
    ////////////////////////////////
    
    public void solve()
    {
      int n=sc.ni();
      Long[] arr=new Long[n];
      HashMap<Long,Integer[]> map=new HashMap();
      for(int i=0;i<n;i++)
        {
            arr[i]=sc.nlo();
            map.put(arr[i],new Integer[33]);
            Arrays.fill(map.get(arr[i]),0);
        }
      Arrays.sort(arr);
      int max=1;
      for(int i=n-1;i>=0;i--)
      {
        for(int j=0;j<33;j++)
        {
            long x=arr[i]+(1L<<j);
            if(map.containsKey(x))
            {
                Integer[] a,b;
                a=map.get(arr[i]);
                b=map.get(x);
                a[j]=b[j]+1;
                a[j]=Math.min(a[j],2);
                max=Math.max(max,a[j]);
            }
        }
      }
      for(int i=0;i<n;i++)
      {
          Integer[] a=map.get(arr[i]);
          for(int j=0;j<33;j++)
          {
              if(a[j]==max)
              {
                  a[j]++;
                  pw.println(a[j]);
                  pw.print(arr[i]);
                  long val=arr[i];
                  while(map.containsKey(val+(1L<<j))&&(a[j]-->1))
                  {
                      val=val+(1L<<j);
                      pw.print(" "+val);
                  }
                  pw.println("");
                  return;
              }
          }
      }
      pw.println(1+"\n"+arr[n-1]);
    }
}