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: Wrong Answer)

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main()
{
	int s=0,a[30],sum=0;
	memset(a,0,sizeof(a));
	string sh;
	cin>>sh;
	for(int i=0;i<=sh.length();i++)
	a[sh[i]-97]++;
	for(int i=0;i<26;i++)
		if(i!=23&&a[i]%2==1)s++;//判断是否需要x的加入 
	if(s>1)sum=-1;
	else
	{
		string t=sh;
		int len=sh.length();
		for(int i=0;i<=len;i++)
		{
			if(t[i]=='x')
			{
				t.erase(i,1);i--;
			}
		}
		//cout<<t<<" "<<t.length()/2<<endl;
		for(int i=0,j=t.length()-1;i<=t.length()/2;i++,j--)
			if(t[i]!=t[j])
			{
				//cout<<t[i]<<"*"<<t[j]<<endl;
				sum=-1;break;
			}
		
		for(int i=0,j=sh.length()-1,k=0;k<=t.length()/2&&i<=j&&sum!=-1;)
		{
			if(sh[i]=='x'&&sh[j]!='x')
			{
				sum++;i++;//cout<<sh[i]<<"*"<<sh[j]<<endl;
			}
			else if(sh[i]!='x'&&sh[j]=='x')
			{
				sum++;j--;//cout<<sh[i]<<"**"<<sh[j]<<endl;
			}
			else if(sh[i]==sh[j])
			{
				i++;j--;k++;//cout<<sh[i]<<"***"<<sh[j]<<endl;
			}
		}
	}
	printf("%d\n",sum);
	return 0;
}