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;
import java.util.Stack;

class Main {

    public static void main(String[] args) {
        new Main().compute();
    }

    void compute() {
        Scanner sc = new Scanner(System.in);
        String S = sc.next();
        String removedX = "";
        int N = S.length();
        boolean[] isX = new boolean[N];
        for (int i = 0; i < N; i++) {/*
            if (S.charAt(i) == 'x') {
                isX[i] = true;
            } else {
                removedX += S.charAt(i);
            }*/
        }
        int rN = removedX.length();
        for (int i = 0; i < rN / 2; i++) {
            if (removedX.charAt(i) != removedX.charAt(rN - i - 1)) {
                System.out.println(-1);
                return;
            }
        }
        int ans = 0;
        int head = 0;
        int foot = N - 1;
        while (foot - head > 0) {
            if (isX[head] == isX[foot]) {
                head++;
                foot--;
            } else if (isX[head] && !isX[foot]) {
                head++;
                ans++;
            } else {
                foot--;
                ans++;
            }
        }
        System.out.println(ans);
    }
}