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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

bool isPalindrome(string s){
	int len=s.length();
	int l=0,r=len-1;
	while(l<r){
		if(s[l]!=s[r]) return false;
		l++,r--;
	}
	return true;
}

int solve(string s, int l, int r){
	if(l>=r) return 0;
	else if(s[l]==s[r]) return solve(s,l+1,r-1);
	else if(s[l]=='x') return 1+solve(s,l+1,r);
	else if(s[r]=='x') return 1+solve(s,l,r-1);
}

int main(){
	string s;
	cin >> s;
	int len=s.length();
	string t="";
	for(int i=0;i<len;i++) if(s[i]!='x') t+=s[i];
	if(!isPalindrome(t)){
		cout << "-1\n";
		return 0;
	}
	printf("%d\n",solve(s,0,len-1));

	return 0;
}