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: Wrong Answer)

import java.util.Scanner;

class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        String s = scanner.next();

        int xCount = 0;
        String newStr = "";

        for (int i = 0; i < s.length(); i++) {
            String tmp = s.substring(i, i + 1);
            if (tmp.equals("x")) {
                xCount++;
            } else {
                newStr += tmp;
            }
        }

        if (check(newStr)) {
            System.out.println(xCount);
        } else {
            System.out.println(-1);
        }
    }

    private static boolean check(String s) {
        int l = s.length();
        if (l % 2 == 1) {
            for (int i = 0; i < (l - 1) / 2; i++) {
                if (!s.substring(i, i + 1).equals(s.substring(l - 1 - i, l - i))) {
                    return false;
                }
            }
        } else {
            for (int i = 0; i < l / 2; i++) {
                if (!s.substring(i, i + 1).equals(s.substring(l - 1 - i, l - i))) {
                    return false;
                }
            }
        }
        return true;
    }
}