You are given a sequence of non-negative integers $A=(a_1,\ldots,a_N)$.

Let us perform the following operation on $A$ just once.

-   Choose a non-negative integer $x$. Then, for every $i=1, \ldots, N$, replace the value of $a_i$ with the bitwise XOR of $a_i$ and $x$.

Let $M$ be the maximum value in $A$ after the operation. Find the minimum possible value of $M$.

What is bitwise XOR? The bitwise XOR of non-negative integers $A$ and $B$, $A \oplus B$, is defined as follows.

-   When $A \oplus B$ is written in binary, the $k$\-th lowest bit ($k \geq 0$) is $1$ if exactly one of the $k$\-th lowest bits of $A$ and $B$ is $1$, and $0$ otherwise.

For instance, $3 \oplus 5 = 6$ (in binary: $011 \oplus 101 = 110$).

The input is given from Standard Input in the following format:

```
$N$
$a_1$ $\ldots$ $a_N$
```

Print the answer.

-   $1 \leq N \leq 1.5 \times 10^5$
-   $0 \leq a_i \lt 2^{30}$
-   All values in the input are integers.

16

If we do the operation with x=2, the sequence becomes (12⊕2,18⊕2,11⊕2)=(14,16,9), where the maximum value M is 16.
We cannot make M smaller than 16, so this value is the answer.