There are `8`

prison cells in a row and each cell is either occupied or vacant.

Each day, whether the cell is occupied or vacant changes according to the following rules:

- If a cell has two adjacent neighbors that are both occupied or both vacant, then the cell becomes occupied.
- Otherwise, it becomes vacant.

**Note** that because the prison is a row, the first and the last cells in the row can't have two adjacent neighbors.

You are given an integer array `cells`

where `cells[i] == 1`

if the `i`

cell is occupied and ^{th}`cells[i] == 0`

if the `i`

cell is vacant, and you are given an integer ^{th}`n`

.

Return the state of the prison after `n`

days (i.e., `n`

such changes described above).

**Example 1:**

Input:cells = [0,1,0,1,1,0,0,1], n = 7Output:[0,0,1,1,0,0,0,0]Explanation:The following table summarizes the state of the prison on each day: Day 0: [0, 1, 0, 1, 1, 0, 0, 1] Day 1: [0, 1, 1, 0, 0, 0, 0, 0] Day 2: [0, 0, 0, 0, 1, 1, 1, 0] Day 3: [0, 1, 1, 0, 0, 1, 0, 0] Day 4: [0, 0, 0, 0, 0, 1, 0, 0] Day 5: [0, 1, 1, 1, 0, 1, 0, 0] Day 6: [0, 0, 1, 0, 1, 1, 0, 0] Day 7: [0, 0, 1, 1, 0, 0, 0, 0]

**Example 2:**

Input:cells = [1,0,0,1,0,0,1,0], n = 1000000000Output:[0,0,1,1,1,1,1,0]

**Constraints:**

`cells.length == 8`

`cells[i]`

is either`0`

or`1`

.`1 <= n <= 10`

^{9}

class Solution {
public int[] prisonAfterNDays(int[] cells, int n) {
}
}