Problem:
Score : 400 points

Problem Statement

We have a string s consisting of lowercase English letters.
Snuke can perform the following operation repeatedly:

Insert a letter x to any position in s of his choice, including the beginning and end of s.

Snuke's objective is to turn s into a palindrome.
Determine whether the objective is achievable. If it is achievable, find the minimum number of operations required.

Notes

A palindrome is a string that reads the same forward and backward.
For example, a, aa, abba and abcba are palindromes, while ab, abab and abcda are not.

Constraints

1 \leq |s| \leq 10^5

s consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

s

Output

If the objective is achievable, print the number of operations required.
If it is not, print -1 instead.

Sample Input 1

xabxa

Sample Output 1

2

One solution is as follows (newly inserted x are shown in bold):

xabxa → xaxbxa → xaxbxax

Sample Input 2

ab

Sample Output 2

-1

No sequence of operations can turn s into a palindrome.

Sample Input 3

a

Sample Output 3

0

s is a palindrome already at the beginning.

Sample Input 4

oxxx

Sample Output 4

3

One solution is as follows:

oxxx → xoxxx → xxoxxx → xxxoxxx

Solution in Rust (1.42.0) (judge verdict: Wrong Answer)

fn main() {
    proconio::input! {
        s: proconio::marker::Chars,
    }
    if s.iter().filter(|x| **x != 'x').collect::<Vec<_>>()
        != s.iter().rev().filter(|x| **x != 'x').collect::<Vec<_>>()
    {
        println!("-1");
    } else {
        let normal = s
            .iter()
            .cloned()
            .collect::<std::collections::VecDeque<char>>();
        let reverse = s
            .iter()
            .cloned()
            .rev()
            .collect::<std::collections::VecDeque<char>>();
        let mut count = 0usize;
        let mut nidx = 0;
        let mut ridx = 0;
        while nidx <= normal.len() / 2 && ridx <= reverse.len() / 2 {
            match (normal[nidx], reverse[ridx]) {
                ('x', 'x') => {
                    nidx += 1;
                    ridx += 1;
                }
                (_, 'x') => {
                    ridx += 1;
                    count += 1;
                }
                ('x', _) => {
                    nidx += 1;
                    count += 1;
                }
                (nc, rc) if nc != rc => unreachable!(),
                (_nc, _rc) => {
                    nidx += 1;
                    ridx += 1;
                }
            }
        }
        println!("{}", count);
    }
}