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.Scanner;

public class Main {

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

		String s = sc.next();

		int headIdx = 0;
		int tailIdx = s.length() - 1;
		int headSkip = 0;
		int tailSkip = 0;
		int ans = 0;
		while (true) {
			while (s.charAt(headIdx) == 'x' && headIdx < s.length() - 1) {
				headIdx++;
				headSkip++;
			}
			while (s.charAt(tailIdx) == 'x' && tailIdx > 0) {
				tailIdx--;
				tailSkip++;
			}
			if (s.charAt(headIdx) != s.charAt(tailIdx)) {
				System.out.print(-1);
				return;
			} else if (headIdx >= tailIdx) {
				ans += Math.abs(headSkip - tailSkip);
				break;
			} else {
				ans += Math.abs(headSkip - tailSkip);
				headSkip = 0;
				tailSkip = 0;
				headIdx++;
				tailIdx--;
			}
		}

		System.out.print(ans);
	}

}