CF 336C(Vasily the Bear and Sequence-贪心-不满足单调性)

C. Vasily the Bear and Sequence

time limit per test

memory limit per test

input

output

a₁, a₂, ..., a_n. Vasily the Bear wants to write out several numbers on a piece of paper so that the beauty of the numbers he wrote out was maximum.

beauty of the written out numbers b₁, b₂, ..., b_k is such maximum non-negative integer v, that number b₁ and b₂ and ... and b_k is divisible by number 2^v without a remainder. If such number v doesn't exist (that is, for any non-negative integer v, number b₁ and b₂and ... and b_k is divisible by 2^v

Tell the bear which numbers he should write out so that the beauty of the written out numbers is maximum. If there are multiple ways to write out the numbers, you need to choose the one where the bear writes out as many numbers as possible.

x and y means applying the bitwise AND operation to numbers x and y. In programming languages C++ and Java this operation is represented by "&", in Pascal — by "and".

Input

n (1 ≤ n ≤ 10⁵). The second line contains n space-separated integers a₁, a₂, ..., a_n (1 ≤ a₁ < a₂ < ... < a_n ≤ 10⁹).

Output

k (k > 0), showing how many numbers to write out. In the second line print k integers b₁, b₂, ..., b_k— the numbers to write out. You are allowed to print numbers b₁, b₂, ..., b_k

Sample test(s)

input

5 1 2 3 4 5

output

2 4 5

input

3 1 2 4

output

1 4

从大往小试，都懂的。

值得一提的是，这不满足单调性。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (100000+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int n,a[MAXN],ans[MAXN];
int main()
{
//  freopen("seq.in","r",stdin);
  cin>>n;
  For(i,n) scanf("%d",&a[i]);
  sort(a+1,a+1+n);
  
  int v=1;
  while (v<a[n]) v<<=1;
  
  while (v)
  {
    int tot=0;
    For(i,n) if (v&a[i]) ans[++tot]=a[i];
    if (!tot) {v>>=1;continue;    }
    int t=ans[1];
    Fork(i,2,tot) t&=ans[i];
    if (t%v==0)
    {
      cout<<tot<<endl;
      For(i,tot-1) cout<<ans[i]<<' ';cout<<ans[tot]<<endl;return 0;
    }
    v>>=1;
  }
  cout<<n<<endl;
  For(i,n-1) cout<<a[i]<<' ';cout<<a[n]<<endl;
  
  
  return 0;
}