Ridhiman challenged Ashish to find the maximum valued subsequence of an array a of size n consisting of positive integers. 

The value of a non-empty subsequence of k elements of a is defined as ∑ 2^i over all integers i ≥ 0 such that at least max(1, k - 2) elements of the subsequence have the i-th bit set in their binary representation (value x has the i-th bit set in its binary representation if ⌊ (x)/(2^i) ⌋ mod 2 is equal to 1). 

Recall that b is a subsequence of a, if b can be obtained by deleting some(possibly zero) elements from a.

Help Ashish find the maximum value he can get by choosing some subsequence of a.

Input

The first line of the input consists of a single integer n (1 ≤ n ≤ 500) — the size of a.

The next line consists of n space-separated integers — the elements of the array (1 ≤ a_i ≤ 10^{18}).

Output

Print a single integer — the maximum value Ashish can get by choosing some subsequence of a.

Examples

Input


3
2 1 3


Output


3

Input


3
3 1 4


Output


7

Input


1
1


Output


1

Input


4
7 7 1 1


Output


7

Note

For the first test case, Ashish can pick the subsequence \{{2, 3}\} of size 2. The binary representation of 2 is 10 and that of 3 is 11. Since max(k - 2, 1) is equal to 1, the value of the subsequence is 2^0 + 2^1 (both 2 and 3 have 1-st bit set in their binary representation and 3 has 0-th bit set in its binary representation). Note that he could also pick the subsequence \{{3\}} or \{{2, 1, 3\}}.

For the second test case, Ashish can pick the subsequence \{{3, 4\}} with value 7.

For the third test case, Ashish can pick the subsequence \{{1\}} with value 1.

For the fourth test case, Ashish can pick the subsequence \{{7, 7\}} with value 7.
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush, nsmallest
from math import ceil, floor, gcd, fabs, factorial, fmod, sqrt, inf
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
# sys.setrecursionlimit(pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]


n = int(data())
arr = l()
answer = arr[0]
if len(arr) == 2:
    answer = arr[0] | arr[1]
for i in range(n-2):
    for j in range(i+1, n-1):
        for k in range(j+1, n):
            answer = max(answer, arr[i] | arr[j] | arr[k])
out(answer)
