# Codeforces Round #486 (Div. 3) D. Points and Powers of Two

xiaoxiao2021-02-28  6

D. Points and Powers of Two time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output

There are nn distinct points on a coordinate line, the coordinate of ii-th point equals to xixi. 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 xi1,xi2,,ximxi1,xi2,…,xim such that for each pair xijxijxikxik it is true that |xijxik|=2d|xij−xik|=2d where dd is some non-negative integer number (not necessarily the same for each pair of points).

Input

The first line contains one integer nn (1n21051≤n≤2⋅105) — the number of points.

The second line contains nn pairwise distinct integers x1,x2,,xnx1,x2,…,xn (109xi109−109≤xi≤109) — the coordinates of points.

Output

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

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

If there are multiple answers, print any of them.

Examples input Copy 6 3 5 4 7 10 12 output Copy 3 7 3 5 input Copy 5 -1 2 5 8 11 output Copy 1 8 Note

In the first example the answer is [7,3,5][7,3,5]. Note, that |73|=4=22|7−3|=4=22|75|=2=21|7−5|=2=21 and |35|=2=21|3−5|=2=21. You can't find a subset having more points satisfying the required property.

#include <bits/stdc++.h> using namespace std; set<long long> s; set<long long>::iterator iter; vector<long long> v; int n; int main() { cin>>n; for(int i=0;i<n;i++) { long long t; cin>>t; s.insert(t); } for(iter=s.begin();iter!=s.end();iter++) { for(int i=0;i<33;i++) { //cout<<*iter-1<<i<<" "<<*iter-(1<<i)<<endl; int flag1 = s.count(*iter-(1<<i)); int flag2 = s.count(*iter+(1<<i)); if(flag1&&flag2) { cout<<3<<endl; cout<<(*iter-(1<<i))<<" "<<*iter<<" "<<(*iter+(1<<i))<<endl; return 0; } else if(flag1) { if(!v.size()) { v.push_back(*iter); v.push_back(*iter-(1<<i)); } } else if(flag2) { if(!v.size()) { v.push_back(*iter); v.push_back(*iter+(1<<i)); } } } } if(v.size()) { cout<<2<<endl; cout<<v[0]<<" "<<v[1]<<endl; } else cout<<1<<endl<<*s.begin()<<endl; }