Score : $600$ points

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

0110 | 0101 = 0111

Constraints

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

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

3
0 1 3
0 2 3

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.

5
1 6 1 4 3
0 6 4 0 1

23

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