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<bits/stdc++.h>

using namespace std;
using ll = long long;
using ld = long double;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define reps(i,s,n) for(int i=(int)(s);i<(int)(n);i++)
const ll mod = 1e9 + 7;
const ll INF = 1e9;
int main() {
	cin.sync_with_stdio(false);
	string s;
	cin >> s;
	string t1, t2;
	int cnt = 0;
	bool flag1 = true;
	rep(i, s.size()) {
		if (s[i] != s[s.size() - 1- i]) {
			flag1 = false;
		}
	}
	if (flag1) {
		cout << 0 << endl;
		return 0;
	}
	rep(i, s.size()) {
		if (s[i] != 'x')t1.push_back(s[i]);
	}
	rep(i, t1.size())t2.push_back(t1[t1.size() - 1 - i]);
	rep(i, t1.size()) {
		if (t1[i] != t2[i]) {
			cout << -1 << endl;
			return 0;
		}
	}
	int len = 0;
	bool flag = true;
	rep(i, s.size()) {
		if (s[i] == 'x')cnt++;
		else len++;
		if (flag&&t1.size() % 2 == 0 && s[len] == 'x') {
			cnt--;
			flag = false;
		}
	}
	cout << cnt << endl;
	return 0;
}