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 (Clang 3.8.0) (judge verdict: Accepted)

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cfloat>
#include <cstring>
#include <map>
#include <utility>
#include <set>
#include <iostream>
#include <memory>
#include <string>
#include <vector>
#include <list>
#include <algorithm>
#include <functional>
#include <sstream>
#include <complex>
#include <stack>
#include <queue>
#include <unordered_set>
#include <unordered_map>
#include <array>
#include <cassert>
#include <bitset>
using namespace std;
using LL = long long;

int main() {
	string S,T;
	cin >> S;
	vector<int>kug(1, 0);
	for (int i = 0; i < S.size(); ++i) {
		if (S[i] != 'x') {
			T.push_back(S[i]);
			kug.push_back(0);
		}
		else {
			kug.back()++;
		}
	}
	for (int i = 0; i < T.size(); ++i) {
		if (T[i] != T[T.size() - 1 - i]) {
			cout << "-1\n";
			return 0;
		}
	}
	int ans = 0;
	for (int i = 0; i < kug.size(); ++i) {
		ans += abs(kug[i] - kug[kug.size() - 1 - i]);
		
	}
	cout << ans / 2 << endl;
	return 0;
}