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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
#define rep(i,n) for (int i=0;i<(int)(n);i++)
#define codefor int test;scanf("%d",&test);while(test--)
#define INT(...) int __VA_ARGS__;in(__VA_ARGS__)
#define LL(...) ll __VA_ARGS__;in(__VA_ARGS__)
#define yn(ans) if(ans)printf("Yes\n");else printf("No\n")
#define YN(ans) if(ans)printf("YES\n");else printf("NO\n")
#define vector1d(type,name,...) vector<type>name(__VA_ARGS__)
#define vector2d(type,name,h,...) vector<vector<type>>name(h,vector<type>(__VA_ARGS__))
#define umap unordered_map
#define uset unordered_set
using namespace std;
using ll = long long;
const int MOD=1000000007;
//入力系
void scan(int& a){scanf("%d",&a);}
void scan(long long& a){scanf("%lld",&a);}
template<class T> void scan(T& a){cin>>a;}
template<class T> void scan(vector<T>& vec){for(auto&& it:vec)scan(it);}
void in(){}
template <class Head, class... Tail> void in(Head& head, Tail&... tail){scan(head);in(tail...);}
//出力系
void print(const int& a){printf("%d",a);}
void print(const long long& a){printf("%lld",a);}
void print(const double& a){printf("%.15lf",a);}
template<class T> void print(const T& a){cout<<a;}
template<class T> void print(const vector<T>& vec){if(vec.empty())return;print(vec[0]);for(auto it=vec.begin();++it!= vec.end();){putchar(' ');print(*it);}}
void out(){putchar('\n');}
template<class T> void out(const T& t){print(t);putchar('\n');}
template <class Head, class... Tail> void out(const Head& head,const Tail&... tail){print(head);putchar(' ');out(tail...);}
//デバッグ系
template<class T> void dprint(const T& a){cerr<<a;}
template<class T> void dprint(const vector<T>& vec){if(vec.empty())return;cerr<<vec[0];for(auto it=vec.begin();++it!= vec.end();){cerr<<" "<<*it;}}
void debug(){cerr<<endl;}
template<class T> void debug(const T& t){dprint(t);cerr<<endl;}
template <class Head, class... Tail> void debug(const Head& head, const Tail&... tail){dprint(head);cerr<<" ";debug(tail...);}
ll intpow(ll a, ll b){ ll ans = 1; while(b){ if(b & 1) ans *= a; a *= a; b /= 2; } return ans; }
ll modpow(ll a, ll b, ll p){ ll ans = 1; while(b){ if(b & 1) (ans *= a) %= p; (a *= a) %= p; b /= 2; } return ans; }
ll updivide(ll a,ll b){if(b==1) return a;else if(a%b==0) return a/b;else return (a/b)+1;}

int main(){
    string s;
    cin>>s;
    int x1=0,x2=0,n=0;
    vector<pair<char,int>> t;
    rep(i,s.length()){
        if(s[i]!='x'){
            n++;
            t.push_back(make_pair(s[i],i));
        }
    }
    if(n==0){
        out(s.length()%2==0);
        return 0;
    }
    int counter=0;
    for(int i=0;i<n;i++){
        counter+=(t[i].first==t[n-1-i].first);
    }
    if(counter!=n){
        out(-1);
        return 0;
    }
    int ans=0,l=0,r=0;
    for(int i=0;i<s.length()/2;i++){
        if(s[i]=='x'&&s[s.length()-1-i]!='x'){
            ans++;
            s.push_back('x');
        }
        else if(s[i]!='x'&&s[s.length()-1-i]=='x'){
            ans++;
            s.insert(s.begin()+i,'x');
            //i++;
        }
    }
    //cout<<s<<endl;
    ans+=(n%2==0);
    out(ans);
}