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>
using namespace std;

int main(){
    string S; cin >> S;
    list<char> Slist, Slist_rev;
    for( int i = 0; i < S.size(); i++ ){
        Slist.emplace_front(S[i]);
        Slist_rev.emplace_back(S[i]);
    }
    auto itr = Slist.begin();
    auto itr_r = Slist_rev.begin();
    int ans = 0;
    while( itr != Slist.end() || itr_r != Slist_rev.end() ){
        if( *itr == *itr_r ){
            itr++;
            itr_r++;
        } else if ( *itr == 'x' ){
            Slist_rev.insert(itr_r,'x');
            itr_r--;
            ans++;
        } else if ( *itr_r == 'x' ){
            Slist.insert(itr,'x');
            itr--;
            ans++;
        } else {
            cout << -1 << endl;
            return 0;
        }
    }
    cout << ans/2 << endl;
}