1582F1 · Korney Korneevich and XOR (easy version)

Problem: This is an easier version of the problem with smaller constraints.

Korney Korneevich dag up an array $a$ of length $n$. Korney Korneevich has recently read about the operation bitwise XOR, so he wished to experiment with it. For this purpose, he decided to find all integers $x \ge 0$ such that there exists an increasing subsequence of the array $a$, in which the bitwise XOR of numbers is equal to $x$.

It didn't take a long time for Korney Korneevich to find all such $x$, and he wants to check his result. That's why he asked you to solve this problem!

A sequence $s$ is a subsequence of a sequence $b$ if $s$ can be obtained from $b$ by deletion of several (possibly, zero or all) elements.

A sequence $s_1, s_2, \ldots , s_m$ is called increasing if $s_1 < s_2 < \ldots < s_m$.

Input Format: The first line contains a single integer $n$ ($1 \le n \le 10^5$).

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 500$) — the elements of the array $a$.

Output Format: In the first line print a single integer $k$ — the number of found $x$ values.

In the second line print $k$ integers in increasing order $x_1, x_2, \ldots x_k$ ($0 \le x_1 < \ldots < x_k$) — found $x$ values.

Note: In the first test case:

• To get value $x = 0$ it is possible to choose and empty subsequence
• To get value $x = 2$ it is possible to choose a subsequence $[2]$
• To get value $x = 4$ it is possible to choose a subsequence $[4]$
• To get value $x = 6$ it is possible to choose a subsequence $[2, 4]$