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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;

int main(){
	string s;
	cin >> s;
	int i = 0, j = s.length() - 1;
	int cnt = 0;
	while(i <= j){
		while(s[i] == 'x' || s[j] == 'x'){
			if(s[i] == 'x' && s[j] != 'x'){
				cnt++;
				i++;
			}else if(s[i] != 'x' && s[j] == 'x'){
				cnt++;
				j--;
			}
		}
		if(s[i] != s[j]){
			printf("-1\n");
			return 0;
		}
		i++; j--;
	}
	printf("%d\n", cnt);
	return 0;
}