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::str::FromStr;
use std::fmt::Debug;
use std::cmp;
#[allow(dead_code)]
fn read_as_vec<T>() -> Vec<T>
where T: FromStr, <T as FromStr>::Err : Debug{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().split_whitespace().map(|c| T::from_str(c).unwrap()).collect()

}

#[allow(dead_code)]
fn read_as_string() -> String{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}

fn is_palind(s: &[char]) -> bool{

    for i in 0..s.len()/2{
        if s[i] != s[s.len()-1-i]{
            return false;
        }
    }
    return true;
}

fn main(){
    let s = read_as_string();
    let sv = s.chars().collect::<Vec<char>>();
    if !s.contains('x'){
        if is_palind(&sv){
            println!("0");
            return;
        }else{
            println!("-1");
            return;
        }
    }else{
        let mut ans: u64 = 0;
        let (mut left, mut right) = (0, sv.len()-1);

        while left < right{
            if sv[left] == sv[right]{
                left += 1;
                right -= 1;
            }else if sv[left] == 'x'{
                left += 1;
                ans += 1;
            }else if sv[right] == 'x'{
                right -= 1;
                ans += 1;
            }else{
                println!("-1");
                return;
            }
        }
        println!("{}", ans);
    }
}