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: Time Limit Exceeded)

#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
#include<cmath>
#include<climits>
#include<string>
#include<set>
#include<numeric>
#include<map>
#include<iostream>
using namespace std;
#define rep(i,n) for(int i = 0;i<((int)(n));i++)
#define reg(i,a,b) for(int i = ((int)(a));i<=((int)(b));i++)
#define irep(i,n) for(int i = ((int)(n)-1);i>=0;i--)
#define ireg(i,a,b) for(int i = ((int)(b));i>=((int)(a));i--)
typedef long long ll;
typedef pair<ll, ll> mp;

/*
AC
外側からxを書いていく．
＝外のxを消していく
*/

int main(void){
	bool ok=true;
	ll ans=0,l=0,r;
	string s;
	cin>>s;
	r=s.size()-1;
	while(l<r){
		if(s[l]=='x'){
			if(s[r]!='x'){
				l++;
				ans++;//外側のxを消していく
			}else{
				//どっちもxなので何もしなくて良い
			}
		}else{
			if(s[r]!='x'){
				if(s[l]==s[r]){
					l++;
					r--;
					//同じなので無問題
				}else{
					//解消不可能
					ok=false;
					break;
				}
			}else{
				r--;
				ans++;//外側のxを消していく
			}
		}
	}
	cout<<(ok?ans:-1)<<endl;
	return 0;
}