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 <algorithm>
#include <iostream>
#include <numeric>
#include <string>
#include <vector>

using namespace std;
vector<int> ch(26);

bool isp(string s) {
  int sl = s.length();
  for (int i = 0; i < sl / 2; i++) {
    if (s[i] != s[sl - 1 - i]) {
      return false;
    }
  }
  return true;
}

int main() {
  string s;
  cin >> s;
  int ans = 0;
  int len = s.length();

  vector <int> f;
  vector <int> l;
  
  //cerr << scut << endl;


  int si = 0;
  int ei = len-1;
  while (true) {
    int fx = 0;
    int lx = 0;
    while(s[si] == 'x'){
      si++;
      fx++;
    }
    while(s[ei] == 'x'){
      ei--;
      lx++;
    }

    //cerr << fx << lx << endl;

    f.push_back(fx);
    l.push_back(lx);

    if(s[si] != s[ei]){
      
          cout << -1 << endl;
          return 0;
          }
    if(ei - si <= 1){
      break;
    }

    si++;
    ei--;



  }

  for(int i = 0; i < f.size(); i++){
    ans += abs(f[i] - l[i]);
  }

  cout << ans << endl;

  return 0;
}