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)

#define  _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <algorithm>
#include <utility>
#include <functional>
#include <cstring>
#include <queue>
#include <stack>
#include <math.h>
#include <iterator>
#include <vector>
#include <string>
#include <set>
#include <math.h>
#include <iostream>
#include <random>
#include<map>
#include <iomanip>
#include <time.h>
#include <stdlib.h>
#include <list>
#include <typeinfo>
#include <list>
#include <set>
#include <cassert>
#include<fstream>
#include <unordered_map>
#include <cstdlib>
using namespace std;
#define Ma_PI 3.141592653589793
#define eps 0.00000000000000000000000001
#define LONG_INF 3000000000000000000
#define GOLD 1.61803398874989484820458
#define MAX_MOD 1000000007
#define REP(i,n) for(long long i = 0;i < n;++i)                                                                             
#define seg_size 524288
int main() {
	string s;
	cin >> s;
	string wow;
	for (int i = 0;i < s.length();++i) {
		if (s[i] != 'x') {
			wow.push_back(s[i]);
		}
	}
	if (wow.length() != 0) {
		for (int i = 0;i < wow.size() / 2;++i) {
			if (wow[i] != wow[wow.size() - i-1]) {
				cout << "-1" << endl;
				return 0;
			}
		}
	}
	//can do
	long long cnt = 0;
	vector<long long> geko;
	int now_itr = 0;
	for (int i = 0;i < s.length();++i) {
		if (now_itr != wow.size()&&s[i] == wow[now_itr]) {
			geko.push_back(cnt);
			now_itr++;
			cnt = 0;
		}
		else {
			cnt++;
		}
	}
	geko.push_back(cnt);
	long long ans = 0;
	for (int i = 0;i < geko.size() / 2;++i) {
		ans += abs(geko[i] - geko[geko.size()-i-1]);
	}
	cout << ans << endl;
	return 0;
}