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 Common Lisp (SBCL 1.1.14) (judge verdict: Runtime Error)

#include <iostream>
#include <list>
#include <string>
#include <algorithm>

using namespace std;

int main(){
    string input;
    cin >> input;
    list<char> list_char((int)input.size());
    copy(input.begin(), input.end(), list_char.begin());
    list<char>::iterator forw = list_char.begin(), backw = list_char.end();
    int sum = 0;
    backw--;
    for(int i = 0; i < (list_char.size() + 1) / 2; i++){
        if(*forw == *backw){
            forw++;
            backw--;
            continue;
        }else if(*backw == 'x'){
            forw = list_char.insert(forw, 'x');
        }else if(*forw == 'x'){
            backw = list_char.insert(next(backw), 'x');
        }else{
            cout << -1 << endl;
            return 0;
        }
        sum++;
        forw++;
        backw--;
    }
    
    //for_each(list_char.begin(), list_char.end(), [](char x){ cout << x; });
    //cout << endl;
    //cout << list_char.size() << endl;
    cout << sum << endl;
    return 0;
}