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: Wrong Answer)

#include <stdio.h>
#include <string.h>
int main(int argc, char *argv[])
{
	char a[1000001],b[1000001];
	int x,j,i;
	while(scanf("%s",&a) != EOF)
	{
		j = 0;
		x = 0;
		for( i = 0; i < strlen(a); i++ )
		{
			if(a[i] != 'x')
			{
				b[j] = a[i];
				j++;
			}
			else x++;
		}
		
		
		
		for( i = 0; i < strlen(a)/2; i++ )
		{
			if( a[i] != b[strlen(a) - 1 - i])break;
		}
		if( i < strlen(a)/2){
			for( i = 0; i < j/2; i++ )
		{
			if( b[i] != b[j - 1 - i])break;
		}
		if( i < j/2)printf("-1\n");
		else printf("%d\n",x);
		}
		else printf("0\n");
		
	} 
	return 0;
}