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>

using namespace std;

#define inf (ll)1e15
#define ll long long
#define all(x) x.begin(), x.end()
#define x first
#define y second

typedef pair<int, int> Point;

const int mod = (int) 1e9 + 7;
const int mod2 = 998244353;
const int MAX_NODES = 1000;

struct Trie{

	int sz;
	int t[MAX_NODES][26], cnt[MAX_NODES][26];

	Trie(){
		memset(t, 0xff, sizeof t);
		memset(cnt, 0, sizeof cnt);
		sz = 0;
	}

	void insert(string s){
		int node = 0;
		for(int i = 0; i < s.size(); ++i){
			if(t[node][s[i] - 'a'] == -1)
				t[node][s[i] - 'a'] = ++sz;
			++cnt[node][s[i] - 'a'];
			node = t[node][s[i] - 'a'];
		}
	}

	int find(string s){
		int node = 0;
		int res = 0;
		for(int i = 0; i < s.size(); ++i){
			res = cnt[node][s[i] - 'a'];
			node = t[node][s[i] - 'a'];
		}
		return res;	
	}

}trie;

int main(){

    cin.tie(0);
    cin.sync_with_stdio(0);

    #ifdef Cyborg101
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #endif

 	string s;
 	cin >> s;

 	int n = s.size();

 	// auto chk = [](string x){
 			
 	// 	int sz = x.size();
 	// 	for(int i = 0; i < sz / 2; ++i){
 	// 		if(x[i] != x[sz - i - 1])
 	// 			return false;
 	// 	}
 	// 	return true;
 	// };
 	

 	// if(chk(s)){
 	// 	cout << 0 << endl;
 	// 	return 0;
 	// }

 	// string pure;

 	int res = 0;

 	int l = 0, r = n - 1;

 	while(l < r){

 		if(s[l] == s[r]) l++, r--;
 		else if(s[l] == 'x') l++, res++;
 		else if(s[r] == 'x') r--, res++;
 		else break;
 	}

 	if(l == r){
 		cout << res << endl;
 	}
 	else{
 		cout << -1 << endl;
 	}
    return 0;
}