### 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 $A'_i=A_{(i+1)\% N}$, 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.

8

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

23

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