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 rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;
const int INT_INF = 2147483647;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

bool check(vector<char> c){
  if(c.size()%2==0)return 0;
  else{
    bool res = 1;
    rep(i,c.size()/2)if(c[i]!=c[c.size()-i-1])res=0;
    return res;
  }
}

int main(){
  string s;
  cin >> s;
  vector<char> c;
  rep(i,s.size())if(s[i]!='x')c.push_back(s[i]);
  //cout << check(c) << endl;
  if(check(c)==0)cout << -1 << endl;
  else{
    int left = 0;
    int right = s.size()-1;
    int ans = 0;
    while(left!=right){
      //cout << left << ' ' << right << endl;
      if(s[left]==s[right])left++,right--;
      else if(s[left]=='x')ans++,left++;
      else ans++,right--;
    }
    cout << ans << endl;
  }
  return 0;
}