Problem Statement

There are $length-N sequences A=(A_0,A_1,\ldots,A_{N-1})$ and $B=(B_0,B_1,\ldots,B_{N-1})$.
Takahashi may perform the following operation on A any number of times (possibly zero):

• apply a left cyclic shift to the sequence A. In other words, replace A with A' defined by A'i=A, where x\% N denotes the remainder when x is divided by N.

Takahashi's objective is to maximize $\displaystyle\sum_{i=0}^{N-1} (A_i|B_i)$, where x|y denotes the bitwise logical sum (bitwise OR) of x and y.
Find the maximum possible $\displaystyle\sum_{i=0}^{N-1} (A_i|B_i)$.

What is the bitwise logical sum (bitwise OR)?
The logical sum (or the OR operation) is an operation on two one-bit integers (0 or 1) defined by the table below.
The bitwise logical sum (bitwise OR) is an operation of applying the logical sum bitwise.

x
y
x|y

0
0
0

0
1
1

1
0
1

1
1
1

The logical sum yields 1 if at least one of the bits x and y is 1.
Conversely, it yields 0 only if both of them are 0.
Example
0110 | 0101 = 0111

Input

The input is given from Standard Input in the following format:
N
$A_0 A_1 \ldots A_{N-1}$
$B_0 B_1 \ldots B_{N-1}$

Output

Print the maximum possible $\displaystyle\sum_{i=0}^{N-1} (A_i|B_i)$.

Constraints

• $2 \leq N \leq 5\times 10^5$
• $0\leq A_i,B_i \leq 31$
• All values in the input are integers.

Sample Input 1

3
0 1 3
0 2 3

Sample Output 1

8

If Takahashi does not perform the operation, A remains (0,1,3), and we have $\displaystyle\sum_{i=0}^{N-1} (A_i|B_i)=(0|0)+(1|2)+(3|3)=0+3+3=6$;
if he performs the operation once, making A=(1,3,0), we have $\displaystyle\sum_{i=0}^{N-1} (A_i|B_i)=(1|0)+(3|2)+(0|3)=1+3+3=7$; and
if he performs the operation twice, making A=(3,0,1), we have $\displaystyle\sum_{i=0}^{N-1} (A_i|B_i)=(3|0)+(0|2)+(1|3)=3+2+3=8$.
If he performs the operation three or more times, A becomes one of the sequences above, so the maximum possible $\displaystyle\sum_{i=0}^{N-1} (A_i|B_i)$ is 8, which should be printed.

Sample Input 2

5
1 6 1 4 3
0 6 4 0 1

Sample Output 2

23

The value is maximized if he performs the operation three times, making A=(4,3,1,6,1),
where $\displaystyle\sum_{i=0}^{N-1} (A_i|B_i)=(4|0)+(3|6)+(1|4)+(6|0)+(1|1)=4+7+5+6+1=23$.

见题面 Input 部分

见题面 Output 部分