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: Accepted)

#include <bits/stdc++.h>
#define pb push_back
#define fi first
#define se second
#define lim (int) (1e5+5)
using namespace std;
typedef pair<int,int> ii;
typedef string string;
typedef long long int lli;
string s;
int ans;
int main()
{
	cin>>s;
	int len=s.length();
	int a=0,b=len-1;
	while(a!=b && a<b)
	{
		if(s[a]==s[b])
		{
			a++;
			b--;
			continue;
		}
		else if(s[a]=='x' && s[b]!='x')
		{
			a++;
			ans++;
			continue;
		}
		else if(s[a]!='x' && s[b]=='x')
		{
			b--;
			ans++;
			continue;
		}
		else
		{
			printf("-1");
			return 0;
		}
	}
	printf("%d",ans);
}