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 "iostream"
#include "climits"
#include "list"
#include "queue"
#include "stack"
#include "set"
#include "functional"
#include "algorithm"
#include "string"
#include "map"
#include "iomanip"

using namespace std;

const long long int MOD = 1000000007;

long long int N, M, K, H, W, L, R;


int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);

	string s;
	cin >> s;
	string t;
	for (auto i : s) {
		if (i != 'x') {
			t += i;
		}
	}
	string u = t;
	reverse(u.begin(), u.end());
	if (t != u) {
		cout << "-1\n";
		return 0;
	}
	vector<int>box;
	int bag = 0;
	for (auto i : s) {
		if (i == 'x') {
			bag++;
		}
		else {
			box.push_back(bag);
			bag = 0;
		}
	}
	box.push_back(bag);
	N = box.size();
	for (int i = 0; i < N / 2; i++) {
		M += abs(box[i] - box[N - i - 1]);
	}
	if (N % 2 && box[N / 2] % 2) {
		M++;
	}
	cout << M << endl;
	return 0;
}