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: Accepted)

#![allow(non_snake_case,unused_imports,dead_code)]
use std::cmp::max;
use std::cmp::min;
use std::str::FromStr;
use std::collections::{HashMap,HashSet,VecDeque,BinaryHeap};
use std::cmp::Ordering;

use proconio::input;
use proconio::marker::{Chars,Usize1,Isize1};

fn comp_f64(a: &f64, b: &f64) -> Ordering {
    if a < b {
        return Ordering::Less;
    } else if a > b {
        return Ordering::Greater;
    }
    Ordering::Equal
}

fn main() {
    input!{
        S:Chars,
    }
    let T = S.iter().filter(|&&s| s!='x').collect::<Vec<_>>();
    let n = T.len();
    for i in 0..n{
        let j = n-1-i;
        if T[i] != T[j]{
            println!("-1");
            return;
        }
    }
    let mut ans = 0;
    let mut i = 0;
    let mut j = S.len()-1;
    while i < j{
        if S[i] == S[j]{
            i += 1;
            j -= 1;
        } else if S[i] == 'x' {
            ans += 1;
            i += 1;
        } else {
            ans += 1;
            j -= 1;
        }
    }
    println!("{}",ans);
}