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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.io.*;

import static java.lang.System.in;

public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        if(!isPalindrome(s.replace("x",""))){
            System.out.println(-1);
            System.exit(0);
        }
        ArrayList<Integer> rec = new ArrayList<>();
        int n = s.length();
        for(int i=0;i<n;i++){
            if(s.charAt(i)!='x') rec.add(i);
        }

        long ans = 0;
        if(rec.size()!=0){
            ans += Math.abs(n-1-rec.get(rec.size()-1)-rec.get(0));
            int flag = 1-rec.size()%2;
            for(int i=0;i<rec.size()/2-flag;i++){
                int left = rec.get(i+1)-rec.get(i);
                int right = rec.get(rec.size()-1-i)-rec.get(rec.size()-2-i);
                ans += Math.abs(left-right);
            }
        }
        System.out.println(ans);
    }
    static boolean isPalindrome(String s){
        if(s.length()==0) return true;
        int n = s.length();
        for(int i=0;i<n;i++){
            if(s.charAt(i)!=s.charAt(n-1-i)) return false;
        }
        return true;
    }
}