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>
#define For(i,a,b) for(i=(a);i<=(b);++i)
#define Forward(i,a,b) for(i=(a);i>=(b);--i)
using namespace std;
template<typename T>inline void read(T &x)
{
	T s=0,f=1;char k=getchar();
	while(!isdigit(k)&&(k^'-'))k=getchar();
	if(!isdigit(k)){f=-1;k=getchar();}
	while(isdigit(k)){s=s*10+(k^48);k=getchar();}
	x=s*f;
}
void file()
{
	#ifndef ONLINE_JUDGE
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
	#endif
}
const int MAXN=700010;
int n,ans;
char a[MAXN];
int main()
{
	scanf("%s",a);
	n=strlen(a)-1;
	int i=0,j=n,ans=0;
	while(i<j)
	{
		int k=0,lj=j;
		while(a[i]=='x'&&i<=n)++i,++k;
		while(a[j]=='x'&&j>=i)--j,--k;
		if(i>j&&j==lj)break;
		ans+=abs(k);
		if(i>j)break;
		if(a[i]!=a[j])
		{
			printf("-1\n");
			return 0;
		}
		++i;--j;
		if(i>j)break;
	}
	printf("%d\n",ans);
	return 0;
}