codeforces631A(暴力枚举)

A. Interview

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

5
1 2 4 3 2
2 3 3 12 1

output

22

input

10
13 2 7 11 8 4 9 8 5 1
5 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, 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 choose l = 1 andr = 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.

//codeforces631A(暴力枚举)
//题目大意：给你两组数a[],b[];已知函数f(a[],l,r)是求a[]数组中下标从l到r的数的按位或运算的结果。
//即f(a[],l,r)=a[l]|a[l+1]|a[l+2]....|a[r].同理f(b[],l,r)也是如此；现在问a[],b[]各有n个数；
//f(a[],l,r)+f(b[],l,r)的最大值.(1<=l<=r<=n) 
#include
#include
int a[1010],b[1010];
int maxk=0,n;
//读取数据函数 
void Input(int a[])   
{ 
    int i;
	for(i=0;imaxk) maxk=u+v;      //更新最大值。 
		}
	  }
    }
	printf("%d\n",maxk);
	return 0;
}