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)

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

#define rep(i, a) for (int i = 0; (i) < (int) (a); (i)++)
#define reps(i, a, b) for (int i = (int) (a); (i) < (int) (b); (i)++)
#define rrep(i, a) for (int i = (int) a-1; (i) >= 0; (i)--)
#define rreps(i, a, b) for (int i = (int) (a)-1; (i) >= (int) (b); (i)--)
#define MP(a, b) make_pair((a), (b))
#define PB(a) push_back((a))
#define all(v) (v).begin(), (v).end()
#define PERM(v) next_permutation(all(v))
#define UNIQUE(v) sort(all(v));(v).erase(unique(all(v)), v.end())
#define CIN(type, x) type x;cin >> x
#define TRUE__  "YES"
#define FALSE__ "NO"
#define PRINT(f) if((f)){cout << (TRUE__) << endl;}else{cout << FALSE__ << endl;}
#define RS resize
#define CINV(v, N) do {\
	v.RS(N);\
	rep(i, N) cin >> v[i];\
} while (0);
#define RCINV(v, N) do {\
	v.RS(N);\
	rrep(i, N) cin >> v[i];\
} while (0);

#define MOD 1000000007

void init();
void solve();

signed main()
{
	init();
	solve();
}

string s;

void init()
{
	cin >> s;
}

void solve()
{
	string t, ti;
	rep(i, s.length()) {
		if (s[i] != 'x') {
			t = t + s[i];
			ti = s[i] + ti;
		}
	}
	if (t != ti) {
		cout << -1 << endl;
		return;
	}
	vector<int> v(1, 0);
	rep(i, s.length()) {
		if (s[i] == 'x') {
			v.back()++;
		} else {
			v.PB(0);
		}
	}
	int si = v.size();
	int res = 0;
	rep(i, si / 2) {
		res += abs(v[i] - v[si - 1 - i]);
	}
	cout << res << endl;
}