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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

//
import java.math.*;
import java.util.*;
import java.io.*;
  
public class Main { 
  
    static BufferedReader in;
    static PrintWriter out = new PrintWriter(System.out);
    static String file = "../in";
    static int test = 10; // 0 for local testing, 1 for std input
    static int inf = 1_000_000, mod = 1_000_000_007;

    void swap(int[]ary, int i, int j)
    {
        int t = ary[i];
        ary[i] = ary[j];
        ary[j] = t;
    }
    
    String[] split() throws Exception
    {
        return in.readLine().split(" ");
    }

    int readInt() throws Exception
    {
        return Integer.valueOf(in.readLine());
    }

    int[] toIntArray() throws Exception
    {
        String[] sp = split();
        int n = sp.length;
        int[] ary = new int[n];
        for(int i = 0; i < n; i++) ary[i] = Integer.valueOf(sp[i]);
        return ary;
    }

    String reverse(String str)
    {
        return new StringBuilder(str).reverse().toString();
    }

    long choose(int n, int a)
    {
        long nf = 1;
        for(int i = 1; i <= a; i++)
        {
            nf *= i;
            nf %= mod;
            nf *= pow(i, mod - 2);
            nf %= mod;
        }
        return nf;
    }

    long pow(int a, int pow)
    {
        return pow(0L + a, pow);
    }
    
    long pow(long a, int pow)
    {
        long res = 1;
        while(pow > 0)
        {
            if(pow % 2 != 0) {
                res *= a;
                res %= mod;
            }
            a = a * a;
            a %= mod;
            pow /= 2;
        }

        return res;
    }    

    public static void main(String[] args) throws Exception
    {
        int _k = Integer.valueOf("1");
        if(test > 0) in = new BufferedReader(new InputStreamReader(System.in));
        else in = new BufferedReader(new FileReader(file));
        if(test < 0) {String[] str = in.readLine().split(" ");}
/***********************************************************************/
/***********************************************************************/
/***********************************************************************/
/***********************************************************************/
        // System.out.println((-100 + 0) / 2);

        new Main().c();
        out.flush();
    }

    void c() throws Exception
    {
        String str = in.readLine();
        // if str without x is palindrome
        // then inserting can make it palin
        String trimmedString = "";
        for(char c : str.toCharArray())
        {
            if(c != 'x') trimmedString += "" + c;
        }

        if(!isPalindrome(trimmedString)) 
        {
            System.out.println(-1);
            return;
        }

        int i = 0, j = str.length() - 1, op = 0;
        while(i < j)
        {
            if(str.charAt(i) == str.charAt(j)) 
            {
                i++;
                j--;
            }
            else
            {
                if(str.charAt(i) == 'x')
                {
                    i++;
                }
                else 
                {
                    j--;
                }
                op++;
            }
        }
        System.out.println(op);
    }

    boolean isPalindrome(String str)
    {
        int i = 0, j = str.length() - 1;
        while(i < j)
        {
            if(str.charAt(i) != str.charAt(j)) return false;
            i++;
            j--;
        }
        return true;
    }

    int find(int[] f , int p)
    {
        int fp = f[p];
        while(fp != f[fp]) fp = f[f[fp]];
        return f[p] = fp;
    }
}