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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
#define MOD 1000000007
#define ll long long
#define sd(a) scanf("%d", &a)
#define sd2(a,b) scanf("%d%d", &a, &b)
#define sd3(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define sld(a) scanf("%lld", &a)
#define sld2(a,b) scanf("%lld%lld", &a, &b)
#define sld3(a, b, c) scanf("%lld%lld%lld", &a, &b, &c)
#define pd(a) printf("%d", a)
#define pld(a) printf("%lld", a)
#define sp printf(" ")
#define nl printf("\n")
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
#define clr(a, val) memset((a), (val), sizeof(a))
#define pb push_back
#define endl "\n"
#define mp(a, b) make_pair(a, b)
#define pii pair<int, int>
#define xx first
#define yy second
#define mx 100001
vector<char> v;
int checkpal(vector<char> &a)
{
	int n=a.size();
	for(int i=0;i<=n/2;i++)
		if(a[i]!=a[n-1-i])
			return 0;
	return 1;
}
vector<char> pass;
int main()
{
	string s;cin>>s;
	for(int i=0;i<s.size();i++)
		if(s[i]!='x')pass.pb(s[i]);
	if(checkpal(pass))
	{
		int ans=0;
		int j=s.size()-1;
		int i=0;
		while(i<j)
		{
			if(s[i]==s[j])
				i++,j--;
			else if(s[i]=='x')
			{ans++;i++;}
			else if(s[j]=='x')
			{ans++;j--;}
		}
		pd(ans);nl;
	}
	else
		printf("-1\n");
}