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)

use proconio::input;
use proconio::marker::Chars;

fn main() {
    input! {
        s: Chars,
    };
    if s.len() == 1 {
        println!("{}", 0);
        return;
    }
    let nx = s
        .iter()
        .filter(|&&c| c != 'x')
        .map(|&c| c)
        .collect::<Vec<char>>();
    let is_ok = {
        let mut nxr = nx.clone();
        nxr.reverse();
        nx == nxr
    };
    if !is_ok {
        println!("{}", -1);
        return;
    }
    if nx.is_empty() {
        println!("{}", 0);
        return;
    }
    let i_m = if nx.len() == 1 {
        s.iter().position(|&c| c == nx[0]).unwrap()
    } else {
        let mut i_m = 0;
        for i in 0..nx.len() / 2 {
            let nx_i = nx[i];
            while nx_i != s[i_m] {
                i_m += 1;
            }
            i_m += 1;
        }
        i_m
    };
    let s_a = s[0..i_m].iter().map(|&c| c).collect::<Vec<char>>();
    let s_b = if nx.len() % 2 == 0 {
        s[i_m..].iter().map(|&c| c).rev().collect::<Vec<char>>()
    } else {
        s[i_m + 1..].iter().map(|&c| c).rev().collect::<Vec<char>>()
    };
    let mut i_a = 0;
    let mut i_b = 0;
    let mut ans = 0;
    while i_a < s_a.len() || i_b < s_b.len() {
        if i_a == s_a.len() {
            ans += 1;
            i_b += 1;
            continue;
        }
        if i_b == s_b.len() {
            ans += 1;
            i_a += 1;
            continue;
        }
        if s_a[i_a] == s_b[i_b] {
            i_a += 1;
            i_b += 1;
            continue;
        } else if s_a[i_a] == 'x' {
            i_a += 1;
            ans += 1;
            continue;
        } else if s_b[i_b] == 'x' {
            i_b += 1;
            ans += 1;
            continue;
        } else {
            unreachable!();
        }
    }
    println!("{}", ans);
}