Bits And Pieces

F. Bits And Pieces

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an array aa of nn integers.

You need to find the maximum value of ai|(aj&ak)ai|(aj&ak) over all triplets (i,j,k)(i,j,k) such that i<j<ki<j<k.

Here && denotes the bitwise AND operation, and || denotes the bitwise OR operation.

Input

The first line of input contains the integer nn (3≤n≤1063≤n≤106), the size of the array aa.

Next line contains nn space separated integers a1a1, a2a2, ..., anan (0≤ai≤2⋅1060≤ai≤2⋅106), representing the elements of the array aa.

Output

Output a single integer, the maximum value of the expression given in the statement.

Examples

input

Copy

3
2 4 6

output

Copy

6

input

Copy

4
2 8 4 7

output

Copy

12

Note

In the first example, the only possible triplet is (1,2,3)(1,2,3). Hence, the answer is 2|(4&6)=62|(4&6)=6.

In the second example, there are 44 possible triplets:

1. (1,2,3)(1,2,3), value of which is 2|(8&4)=22|(8&4)=2.
2. (1,2,4)(1,2,4), value of which is 2|(8&7)=22|(8&7)=2.
3. (1,3,4)(1,3,4), value of which is 2|(4&7)=62|(4&7)=6.
4. (2,3,4)(2,3,4), value of which is 8|(4&7)=128|(4&7)=12.

The maximum value hence is 1212.

 

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int maxn = (1 << 24);
const int M = 23;
int n, cnt[maxn];

inline void update(ll cur, int pos = 0) {
    if (cnt[cur] >= 2)return;
    cnt[cur]++;
    for (register int i = pos; i <= M; ++i) {
        if (cur & (1 << i)) {
            update(cur ^ (1 << i), i);
        }
    }
}

int main() {
    //freopen("1.txt", "r", stdin);
    scanf("%d", &n);
    vector<ll> e(n);
    for (auto &i : e) {
        scanf("%lld", &i);
    }
    //for(register int i=0;i<3;++i)printf("%lld\n",e[i]);
    update(e[n - 1]);
    update(e[n - 2]);
    ll res = 0;

    for (register int i = n - 3; i >= 0; --i) {
        ll head = e[i];
        int nx = 0;
        for (register int j = M; j >= 0; --j) {
            if (!(head & (1 << j)) && cnt[nx | (1 << j)] >= 2) {
                nx |= (1 << j);
            }
            //printf("debug res = %lld\n",res);
        }
        res = max(res, head | nx);

        update(e[i]);
    }
    printf("%lld\n", res);
    return 0;
}