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(void){
	char s[100000],S[100000];
	int len,i,j=0,x=0;
	scanf("%s",&s);
	for(i=0;i<strlen(s);i++){
		if(s[i]=='x') x++;
		else{
			S[j] = s[i];
			j++;
		}
	}
	len = strlen(S);
	for(i=0;i<len/2.0;i++){
		if(S[i]!=S[len-1-i]){
			printf("%d\n",-1);
			return 0;
		}else{
			if(s[0]=='x'||s[strlen(s)-1]=='x') printf("%d\n",len+1-x);
			else printf("%d\n",len-1-x);
		}
	}
	return 0;
}