Interview（位運算）

Interview

Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 631A

Description

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 x_l, x_l + 1, ..., x_r, where x_i 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 a_i (0 ≤ a_i ≤ 10⁹).

The third line contains n integers b_i (0 ≤ b_i ≤ 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.

Sample Input

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

Sample Output

Hint

Bitwise OR of two non-negative integers a and b is the number c = aORb, 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. （或運算）

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.

注；這題主要考查位運算：每一列中，得出最大一個數，這個數的每個二進制位上的1，在這一列數中都能找到一個或多個數在相同的二進制位置上的1與之對應。

之前想着循環除以2得到每個數的2進制位，不過感覺會超時，所以就百度別人的題解，發現直接用或運算很簡便。

My   solution:

/*2016.3.12*/

#include<stdio.h>
int main()
{
	int i,j,n,m,k;
	int a[1100],b[1100];
	while(scanf("%d",&n)==1)
	{
		for(i=1;i<=n;i++)
		scanf("%d",&a[i]);
		for(i=1;i<=n;i++)
		scanf("%d",&b[i]);
		for(i=2;i<=n;i++)
		{
			a[1]|=a[i];//與運算 
			b[1]|=b[i];
		}
		k=a[1]+b[1];
		printf("%d\n",k);	 
	}
	return 0;
}