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.

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).

Print a single integer — the maximum value of sum f(a, l, r) + f(b, l, r) among all possible 1 ≤ l ≤ r ≤ n.

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 1if 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, becausef(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 choosel = 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：给两个序列，要求找到一个区间（l，r）使得在这个区间内序列a的数相互或运算和序列b的数或运算之后和最大，输出最大和

题解：题目中的区间就是一个误导项，因为是或运算，遇见1即为1所以我们把序列中所有数或运算即可

#include<stdio.h>

#include<string.h>

#include<string>

#include<math.h>

#include<algorithm>

#define LL long long

#define PI atan(1.0)*4

#define DD double

#define MAX 300100

#define mod 100

#define dian 1.000000011

#define INF 0x3f3f3f

using namespace std;

char s[MAX];

int a[MAX],b[MAX];

int main()

{

	int n,i,j;

	while(scanf("%d",&n)!=EOF)

	{

		for(i=1;i<=n;i++)

		    scanf("%d",&a[i]);

		for(i=1;i<=n;i++)

		    scanf("%d",&b[i]);

		int x=a[1];

		int y=b[1];

		for(i=2;i<=n;i++)

		{

			x=max(x,(x|a[i]));

			y=max(y,(y|b[i]));

		}

		printf("%d\n",x+y);

	}

	return 0;

}