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: Accepted)

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

int main() {
	int ans=0,flag=0,first=0,last=99999;
	char s[100001];
	scanf("%s", s);
/*回文の先頭と末尾が同じならその2つは無視し除外できる。
sはkeepしfirstとlastの位置で何とかする。*/
	last = strlen(s)-1;
	while (flag == 0) {
		if (s[first] != s[last] && s[first] != 'x' && s[last] != 'x')flag++;
		else if (s[first] == s[last]) { first++; last--; }
		else if (s[first] == 'x' && s[last] != 'x') { first++; ans++; }
		else if (s[first] != 'x' && s[last] == 'x') { last--; ans++; }
		
		if (first >= last) { printf("%d", ans); return 0; }
	}
	printf("-1");
}