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;

typedef long long ll;
typedef vector<long long> VLL;
typedef vector<pair<int,int>> VP;
#define INF (int)(1e9)
#define MAXX 1.1529215e+18
#define inf 999999
#define EPS (1e-7)
#define MOD (1e9+7)
#define rep(i,n) for(int i=0; i<(int)(n);i++)
#define REP(i,n) for(int i=1;i<=(int)(n);i++)
#define FOR(i,k,n) for(int i=(k);i<(int)(n);i++)
#define ALL(a)  a.begin(),a.end()
#define RALL(a) a.begin(),a.end(),greater<int>()
#define ROT(a) a.begin(),a.begin()+1,a.end()
#define RROT(a) a.begin(),a.end()-1,a.end()
#define PB push_back
#define MP make_pair
#define PI acos(-1.0)
template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; }
template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return 1; } return 0; }
template<typename T> T gcd(T a, T b) { if (b == 0) return a; return gcd(b, a % b); }
/*--------------------------------------------*/

int main(){
  string s;
  cin >> s;
  
  s.insert(s.begin(),'a');
  s.insert(s.end(),'a');
  
  int N = s.size();
  
  vector<pair<char,int> > t;
  
  rep(i,N){
    if(s[i] != 'x'){
      t.PB(MP(s[i],i));
    }
  }
  
  int M = t.size();
  
  int cost = 0;
  
  if(M % 2 == 0){
    rep(i,M/2){
      if(t[i].first != t[M-i-1].first){
        cout << -1 << endl;
        return 0;
      }
    }
    
    for(int i = M/2;i > 1;i--){
      int left = t[i-1].second - t[i-2].second-1;
      int right = t[M-i+1].second - t[M-i].second-1;
      cost += abs(left-right);
    }
      
  }else{
    rep(i,(M-1)/2){
      if(t[i].first != t[M-i-1].first){
        cout << -1 << endl;
        return 0;
      }
    }
      
    for(int i = (M-1)/2;i > 0;i--){
      int left = t[i].second - t[i-1].second-1;
      int right = t[M-i].second - t[M-i-1].second-1;
      cost += abs(left-right);
    }
    
  }

  cout << cost << endl;
  
}