Points and Powers of Two CodeForces - 988D转自：https://www.cnblogs.com/xingkongyihao/p/9124635.htmlD. Points and Powers of Two

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我是靠谱客的博主自然烤鸡，最近开发中收集的这篇文章主要介绍Points and Powers of Two CodeForces - 988D转自：https://www.cnblogs.com/xingkongyihao/p/9124635.htmlD. Points and Powers of Two，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

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 ${xi1,xi2,\dots,ximxi1,xi2,\dots,xim such that for each pair xijxij, xikxik it is true that |xij-xik|=2d|xij-xik|=2d where dd is some non-negative integer number (not necessarily the same for each pair of points).}_{{i1,xi2,\dots,ximxi1,xi2,\dots,xim such that for each pair xijxij, xikxik it is true that |xij-xik|=2d|xij-xik|=2d where dd is some non-negative integer number (not necessarily the same for each pair of points).}_{1,xi2,\dots,ximxi1,xi2,\dots,xim such that for each pair xijxij, xikxik it is true that |xij-xik|=2d|xij-xik|=2d where dd is some non-negative integer number (not necessarily the same for each pair of points).}}$

Input

The first line contains one integer $nn (1\leqn\leq2\cdot1051\leqn\leq2\cdot105) — the number of points.$

The second line contains $nn pairwise distinct integers x1,x2,\dots,xnx1,x2,\dots,xn (-109\leqxi\leq109-109\leqxi\leq109) — the coordinates of points.$

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 $[7,3,5][7,3,5]. Note, that |7-3|=4=22|7-3|=4=22, |7-5|=2=21|7-5|=2=21 and |3-5|=2=21|3-5|=2=21. You can't find a subset having more points satisfying the required property.$

题意：选取最大的区间，使的两两之差是2^d ，d是非负整数。

思路：有一个结论，就是这样的区间的元素个数最多为3！好了，知道了这个结论就很好写了。

#include <bits/stdc++.h>
using namespace std;
const int N = 2e5+10;
int a[N], n;
map<int,int> mp;
int main() {
    scanf("%d",&n);
    for(int i = 0; i < n; i ++) {
        scanf("%d",&a[i]);
        mp[a[i]] = 1;
    }
    sort(a,a+n);
    for(int i = 0; i < n; i ++) {
        for(int j = 0; j < 32; j ++) {
            int x = 1 << j;
            if(mp[a[i]+x] && mp[a[i]+2*x]) {
                return 0*printf("3n%d %d %dn",a[i],a[i]+x,a[i]+x*2);
            }
            if(a[i]+x > a[n-1] || a[i]+2*x > a[n-1]) break;
        }
    }
    for(int i = 0; i < n; i ++) {
        for(int j = 0; j < 32; j ++) {
            int x = 1 << j;
            if(mp[a[i]+x]) {
                return 0*printf("2n%d %dn",a[i],a[i]+x);
            }
            if(mp[a[i]+x*2]) {
                return 0*printf("2n%d %dn",a[i],a[i]+2*x);
            }
            if(a[i]+x > a[n-1] || a[i]+2*x > a[n-1]) break;
        }
    }
    printf("1n%dn",a[0]);
    return 0;
}