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;
typedef long long ll;
using P   = pair<ll, ll>;
using mat = vector<vector<ll>>;

void IOS() { ios::sync_with_stdio(false), cin.tie(0); }

const ll INF = 1e16;
const ll MOD = 1e9 + 7;

ll mod(ll a) {
	if (a > 0) return a % MOD;
	if (a % MOD == 0) return 0;
	ll x = -a / MOD + 1;
	a += x * MOD;
	return a % MOD;
}

template <typename T> void dump(T x) { cout << x << endl; }
template <typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val) {
	fill((T *)array, (T *)(array + N), val);
}
// int dx[4]    = {1, 0, -1, 0};
// int dy[4]    = {0, 1, 0, -1};
// ll pas[5001][5002];

ll cnt[26] = {};

int main() {
	IOS();
	string s, t;
	t = "";
	cin >> s;
	int r  = s.size() - 1;
	int l  = 0;
	ll ans = 0;
	while (l < r) {
		if (s[l] == s[r]) {
			l++;
			r--;
		} else if (s[l] == 'x')
			l++, ans++;
		else if (s[r] == 'x')
			r--, ans++;
		else {
			dump(-1);
			return 0;
		}
	}
	dump(ans);
}