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: Runtime Error)

#include <stdio.h>
#include <string.h>

int main(){
	char str[100000], str2[100000];
	scanf("%s", str);
	int len = strlen(str);

	int num=0;
	for(int i=0; str[i]!='\0'; i++){
		if(str[i]!='x'){
			str2[num] = str[i];
			num++;
		}
	}
	str2[num] = '\0';

	//printf("%s\n%s\n", str, str2); //

	int len2=strlen(str2);
	//printf("%d\n", len2); //
	for(int i=0; i<len2/2; i++){
		if(str2[i]!=str2[len2-1-i]){
			printf("-1\n");
			//printf("%d %d\n", i, len2-1-i); //
			//printf("%c %c\n", str2[i], str2[len-1-i]); //
			return 0;
		}
	}

	int center=0;
	int center2=(len2+1)/2;
	for(int i=0; 0<center2; i++){
		center++;
		if(str[i]!='x'){
			center2--;
		}
	}

	int answer=0;
	for(int i=0; i<len; i++){
		if(str[i]=='x'){
			answer++;
		}
	}
	for(int i=0; i<len/2; i++){
		if(str[i]=='x' && str[len-1-i]=='x' && i<=center && len-1-i>center){
			answer -= 2;
		}
	}
	
	printf("%d\n", answer);
	
	return 0;
}