We have $N$ non-negative integers: $A_1, A_2, ..., A_N$.

Consider painting at least one and at most $N-1$ integers among them in red, and painting the rest in blue.

Let the _beauty_ of the painting be the $\text{XOR}$ of the integers painted in red, plus the $\text{XOR}$ of the integers painted in blue.

Find the maximum possible beauty of the painting.

What is $\text{XOR}$?

The bitwise $\text{XOR}$ $x_1 \oplus x_2 \oplus \ldots \oplus x_n$ of $n$ non-negative integers $x_1, x_2, \ldots, x_n$ is defined as follows:

-   When $x_1 \oplus x_2 \oplus \ldots \oplus x_n$ is written in base two, the digit in the $2^k$'s place ($k \geq 0$) is $1$ if the number of integers among $x_1, x_2, \ldots, x_n$ whose binary representations have $1$ in the $2^k$'s place is odd, and $0$ if that count is even.

For example, $3 \oplus 5 = 6$.

Input is given from Standard Input in the following format:

```
$N$
$A_1$ $A_2$ $...$ $A_N$
```

Print the maximum possible beauty of the painting.

-   All values in input are integers.
-   $2 \leq N \leq 10^5$
-   $0 \leq A_i < 2^{60}\ (1 \leq i \leq N)$

12

If we paint $3$, $6$, $5$ in blue, red, blue, respectively, the beauty will be $(6) + (3 \oplus 5) = 12$.

There is no way to paint the integers resulting in greater beauty than $12$, so the answer is $12$.