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<bits/stdc++.h>
#define loop(i, a, b) for(int i = a; i < b; i++) 
#define rep(i, a) loop(i, 0, a)
using namespace std;
const int MOD = 1000000007;
const int inf = 1e9;
using vi = vector <int>;
using vvi = vector <vi>;
using vc = vector <char>;
using vvc = vector <vc>;
using vs = vector <string>;
using vvs = vector <vs>;
//g++ -std==c++14
//setprecision(10)

int main(){
    string s;
    cin >> s;
    int num_x = 0;
    string non_x = "";
    rep(i, s.size()){
        if(s[i] != 'x')
            non_x += s[i];
        else
            num_x++;
    }
    char va = non_x[non_x.size() / 2];
    int ri = 0;
    rep(i, non_x.size() / 2){
        if(non_x[i] != non_x[non_x.size()- 1 - i]){
            cout << -1 << endl;
            return 0;
        }
        if(non_x[i] == va)
            ri++;
    }
    int memo;
    rep(i, s.size()){
        if(s[i] == va)ri--;
        if(ri == 0){
            memo = i;
            break;
        }
    }
    int le = 0;
    ri = 0;
    rep(i,s.size()){
//        cout << s[i] << endl;
        if(s[i] == 'x'){
//            cout << i << endl;
            (i < memo ? le : ri) += 1;
        }
    }
    //cout << ri << " " << le << endl;   
    cout << max(ri, le) << endl;
}