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

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 $n$ 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 $x_{i_1},x_{i_2},\dots ,x_{i_m}$ 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 $n$ ($1\le n\le 2\cdot {10}^5$) — the number of points.

The second line contains $n$ pairwise distinct integers $x_1,x_2,\dots ,x_n$ ($-{10}^9\le x_i\le {10}^9$) — 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

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 $[7,3,5]$. 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.

思路：

可以先写几组数据分析一下怎么构造解。先排序，大小为1和2就不用说了，大小为3的时候，相邻的差值一定相等且为2的幂，这样最大和最小的差值一定也是2的幂。分析大小为4的时候，如果前三个数按照以上规则构造，那么最后一个数一定构造不出来。由此可知大小最大为3。并且贪心策略也有了。这一题用map会超，可以用二分查找。

代码：

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
using namespace std;
#define inf 0x3f3f3f3f
#define MAXN 200005
#define ll long long
int n,ans;
ll a[MAXN],b[10],fac[66];
int main()
{
    scanf("%d",&n);
    fac[0]=1;
    for(int i=1;i<=63;i++)
    {
        fac[i]=fac[i-1]<<1;
    }
    for(int i=0;i<n;i++)
    {
        scanf("%lld",&a[i]);
    }
    sort(a,a+n);
    ans=1;
    b[0]=a[0];
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<=62;j++)
        {
            if(i+2<n && (*lower_bound(a,a+n,a[i]+fac[j]))==a[i]+fac[j] && (*lower_bound(a,a+n,a[i]+fac[j+1]))==a[i]+fac[j+1])
            {
                printf("3\n");
                printf("%lld %lld %lld\n",a[i],a[i]+fac[j],a[i]+fac[j+1]);
                exit(0);
            }
            else if(i+1<n && (*lower_bound(a,a+n,a[i]+fac[j]))==a[i]+fac[j] && ans==1)
            {
                ans=2;
                b[0]=a[i];
                b[1]=a[i]+fac[j];
            }
        }
    }
    printf("%d\n",ans);
    for(int i=0;i<ans;i++)
    {
        if(i) printf(" ");
        printf("%lld",b[i]);
    }
    printf("\n");
	return 0;
}