Blake is a CEO of a large company called "Blake Technologies". He loves his company very much and he thinks that his company should be the best. That is why every candidate needs to pass through the interview that consists of the following problem.
We define function f(x, l, r) as a bitwise OR of integers xl, xl + 1, ..., xr, where xi is the i-th element of the array x. You are given two arrays a and b of length n. You need to determine the maximum value of sum f(a, l, r) + f(b, l, r) among all possible 1 ≤ l ≤ r ≤ n.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 1000) — the length of the arrays.
The second line contains n integers ai (0 ≤ ai ≤ 109).
The third line contains n integers bi (0 ≤ bi ≤ 109).
Output
Print a single integer — the maximum value of sum f(a, l, r) + f(b, l, r) among all possible 1 ≤ l ≤ r ≤ n.
Examples
Input
51 2 4 3 22 3 3 12 1
Output
22
Input
1013 2 7 11 8 4 9 8 5 15 7 18 9 2 3 0 11 8 6
Output
46
Note
Bitwise OR of two non-negative integers a and b is the number c = a OR b, such that each of its digits in binary notation is 1 if and only if at least one of a or b have 1 in the corresponding position in binary notation.
In the first sample, one of the optimal answers is l = 2 and r = 4, because f(a, 2, 4) + f(b, 2, 4) = (2 OR 4 OR 3) + (3 OR 3 OR 12) = 7 + 15 = 22. Other ways to get maximum value is to choose l = 1 and r = 4, l = 1 and r = 5, l = 2 and r = 4, l = 2 and r = 5, l = 3 and r = 4, or l = 3 and r = 5.
In the second sample, the maximum value is obtained for l = 1 and r = 9.
题意（虽然题目很水，不过位运算太弱了，还是写了一下）：
函数f（a, l, r）表达式为：al | al+1 | al+2 | ...... | ar，其中a为一个集合，ai表示集合a中第i个元素；
现有集合a , b （都有n个元素），求f(a, l, r) + f(b, l , r)的最大值；
思路：首先我想到的就是两个for找l和r，因为数据只有1000，肯定是不会超时的；
不过我对位运算并不熟悉，所以先按样例解析跑了一遍。。
发现2 | 4 | 3 = 1 | 2 | 4 | 3 = 2 | 4 | 3 | 2 = 1 | 2 | 4 | 3 | 2 ！！！！很神奇有木有！！！！
由此猜测对任意一个集合的所有子集的所有按位或运算。。最大值等于这个集合本身所有元素间进行位运算。。
样例2也符合这个规律。。。
仔细一想。。不难发现任意两个数进行位或运算后得到的值必然是非减的。。这可以由位或运算的规则证明。。
位或其功能是参与运算的两数各对应的二进位相或。只要对应的二个二进位有一个为1时，结果位就为1。2进制对应位只要有一个为1，其结果就是1.。所有运算结果不可能比原数小。。。
所有这个题目只需要分别把两个集合的元素进行位或运算再相加就可以了啦。。。
代码：
#include <bits/stdc++.h>
#define ll long long
#define MAXN 100000+10
using namespace std;
int main(void)
{
int n, x, y, cnt1=0, cnt2=0; // 注意cnt1， cnt2赋初值0，如果赋值1的话其结果可能变大
cin >> n;
for(int i=0; i<n; i++)
{
cin >> x;
cnt1|=x;
}
for(int j=0; j<n; j++)
{
cin >> y;
cnt2|=y;
}
cout << cnt1+cnt2 << endl;
return 0;
}