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 (1.15.1) (judge verdict: Accepted)

use std::io::Read;
use std::collections::VecDeque;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();

    let answer = solve(&buf);

    println!("{}", answer);
}

fn solve(input: &str) -> String {
    let mut iterator = input.split_whitespace();

    let mut s: VecDeque<char> = iterator.next().unwrap().chars().collect();

    let mut ans = 0;
    while s.len() >= 2 {
        let mut f = ' ';
        let mut b = ' ';
        {
            f = *s.front().unwrap();
            b = *s.back().unwrap();
        }
        if f == b {
            s.pop_front().unwrap();
            s.pop_back().unwrap();
        } else if f == 'x' {
            s.push_back('x');
            ans += 1;
        } else if b == 'x' {
            s.push_front('x');
            ans += 1;
        } else {
            return "-1".to_string();
        }
    }

    return ans.to_string();
}