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)

/* 
	Author: ankrypt
*/
#include <bits/stdc++.h>

using namespace std;

string S;
int cnt = 0;
int flag = 0;

int sizeOfString;

int main()
{
	cin >> S;
	sizeOfString = S.size();
	int i = 0;
	int j = S.size() - 1;

	while(i != j)
	{
		if(S[i] == S[j])
		{
			i++;
			j--;
			if(i > j)
			{
				break;
			}
		}
		else if(S[i] == 'x' && S[i] != S[j])
		{
			if(flag == 0 && S[0] != 'x' && S[sizeOfString - 1] != 'x')
			{
				cnt += 2;
				flag = 2;
			}
			i++;
			cnt++;
		}
		else if(S[j] == 'x' && S[i] != S[j])
		{
			if(flag == 0 && S[0] != 'x' && S[sizeOfString - 1] != 'x')
			{
				cnt += 2;
				flag = 2;
			}
			j--;
			cnt++;
		}
		else if(S[j] != S[i] && S[i] != 'x' && S[j] != 'x')
		{
			flag = 1;
			break;
		}
	}
	if(flag == 1)
	{
		printf("-1\n");
	}
	else
	{
		printf("%d\n", cnt);
	}
	
	return 0;
}
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