C. Vasily the Bear and Sequence

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasily the bear has got a sequence of positive integers 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.

The 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 2v without a remainder. If such number v doesn't exist (that is, for any non-negative integer v, number b1 and b2 and ... and b**k is divisible by 2v without a remainder), the beauty of the written out numbers equals -1.

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.

Here expression 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

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

Output

In the first line print a single integer 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 in any order, but all of them must be distinct. If there are multiple ways to write out the numbers, choose the one with the maximum number of numbers to write out. If there still are multiple ways, you are allowed to print any of them.

Examples

input

Copy

51 2 3 4 5

output

Copy

24 5

input

Copy

31 2 4

output

Copy

14

题意：

给你一个含有n个数的数组a

让你找到一个数组a的子集，满足：子集中的每一个数相与得到的结果x，满足一个最大的k，\(x\) 对$2^k $ 取模为0.

在满足k最大的前提下，子集的集合大小最大。

思路：

枚举k从30 到0 ，2^K 即 为 \(2^{30},2^{29},2^{28} ... 2^{0}\)

对于每一个k，我们让数组a中所有第k位为1的数都相与起来，判断相与后的数值x的lowbit(x)是否为2…^k。

如果x的lowbit(x) 为k，就代表x对 2^k 取模为0，如何题意，且k是满足题意最大的，

集合就为数组a中所有第k位为1的数。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {a %= MOD; if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}

inline void getInt(int* p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int n;
int a[maxn];
std::vector<int> v[50];
std::vector<int> ans2;
int lowbit(int x)
{
    return -x & x;
}
int main()
{
    //freopen("D:\\code\\text\\input.txt","r",stdin);
    //freopen("D:\\code\\text\\output.txt","w",stdout);
    gbtb;
    cin >> n;
    int num = 0;
    repd(i, 1, n)
    {
        cin >> a[i];
    }
    for (int j = 30; j >= 0; j--)
    {
        int num = (1 << j);
        int x = -1;
        repd(i, 1, n)
        {
            if (num & (a[i]))
            {
                if (x == -1)
                {
                    x = a[i];
                } else
                {
                    x &= a[i];
                }
                ans2.push_back(a[i]);
            }
        }
        if (lowbit(x) == num)
        {
            break;
        } else
        {
            ans2.clear();
        }
    }
    cout << sz(ans2) << endl;
    Pv(ans2);

    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}