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 <string>
#include <algorithm>

using namespace std;

int main() {
    string S;
    int num1 = 0;
    int num2 = 0;
    int num3 = 0;
    int tmp = 0;
    int ret = 0;
    int pos = 0;
    cin >> S;
    for (int i = 0; i < S.size(); ++i)
    {
        if (S[i] != 'x')
        {
            tmp = num3;
            num3 = 0;
            for (int j = pos; j < S.size() - i; ++j)
            {
                if (S[S.size() - 1 - j] != 'x')
                {
                    if (S[S.size() - 1 - j] == S[i])
                    {
                        pos = j;
                        break;
                    }
                    else
                    {
                        cout << -1 << endl;
                        return 0;
                    }
                }
                else
                {
                    num2++;
                }
            }
        }
        else
        {
            num1++;
            num3++;
            if (i == S.size() - 1 - pos)
            {
                ret = tmp + (max(num1 - tmp, num2) * 2) - (num1 + num2);
                break;
            }
        }
        if (pos == S.size() - 1 - i || pos == S.size() - i - 2)
        {
            ret = (max(num1, num2) * 2) - (num1 + num2);
            break;
        }
    }
    cout << ret << endl;
    return 0;
}