Problem Statement

You are given a sequence of non-negative integers $A = (A_1, \\ldots, A_N)$. You can do the operation below any number of times (possibly zero).

• Choose an integer $x\\in \\{A_1,\\ldots,A_N\\}$.
• Replace $A_i$ with $A_i\\oplus x$ for every $i$ ($\\oplus$ denotes the bitwise $\\mathrm{XOR}$ operation).

Find the maximum possible value of $\\sum_{i=1}^N A_i$ after your operations.

What is bitwise $\\mathrm{XOR}$?

The bitwise $\\mathrm{XOR}$ of non-negative integers $A$ and $B$, $A \\oplus B$, is defined as follows:

• When $A \\oplus B$ is written in base two, the digit in the $2^k$'s place ($k \\geq 0$) is $1$ if exactly one of $A$ and $B$ is $1$, and $0$ otherwise.

For example, we have $3 \\oplus 5 = 6$ (in base two: $011 \\oplus 101 = 110$).

Constraints

• $1\\leq N \\leq 3\\times 10^{5}$
• $0\\leq A_i < 2^{30}$

Input

Input is given from Standard Input from the following format:

$N$ $A_1$ $\\ldots$ $A_N$

Output

Print the maximum possible value of $\\sum_{i=1}^N A_i$ after your operations.

Sample Input 1

5
1 2 3 4 5

Sample Output 1

19

Here is a sequence of operations that achieves $\\sum_{i=1}^N A_i = 19$.

• Initially, the sequence $A$ is $(1,2,3,4,5)$.
• An operation with $x = 1$ changes it to $(0,3,2,5,4)$.
• An operation with $x = 5$ changes it to $(5,6,7,0,1)$, where $\\sum_{i=1}^N A_i = 5 + 6 + 7 + 0 + 1 = 19$.

Sample Input 2

5
10 10 10 10 10

Sample Output 2

50

Doing zero operations achieves $\\sum_{i=1}^N A_i = 50$.

Sample Input 3

5
3 1 4 1 5

Sample Output 3

18