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 C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <string.h>
int main() {
	char a[100010];
	int ans = 0, len;
	int l = 0, r = 1;
	scanf("%s", a);
	len = strlen(a);
	for (int i = 0; i + l <= len - i - r; i++)
	{
		if (a[i + l] != a[len - i - r]) {
			if (a[i + l] == 'x') {
				i--;
				l++;
				ans++;
			}
			if (a[len - i - r] == 'x') {
				i--;
				r++;
				ans++;
			}
			if (a[i + l] != 'x' && a[len - i - r] != 'x') {
				printf("-1");
				return 0;
			}
		}
	}
	printf("%d", ans);
	return 0;
}