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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace C_Insertingx
{
    class Program
    {
        static void Main(string[] args)
        {

            string strS = Console.ReadLine();

            if (strS.Length == 1) {
                Console.WriteLine(0);
                return;
            }




            string strS2 = strS;
            int intCnt = 0;

            for (int i = 0; i < strS.Length; i++)
            {
                int intHalfL2 = strS2.Length / 2;
                int k = strS2.Length - 1;
                for (int j = 0; j < intHalfL2; j++)
                {
                    string strX = strS2.Substring(j, 1); // 文字列先頭
                    string strY = strS2.Substring(k, 1); // 文字列末尾
                    if (strX != strY)　// 先頭と末尾を比較
                    {
                        if (strX != "x" && strY != "x") { // 先頭も末尾もxではない→回文は作成できない
                            Console.WriteLine(-1);
                            return;
                        }

                        intCnt++;
                        if (strX == "x")
                        {
                            strS2 += "x";
                            break;
                        }

                        if (strY == "x")
                        {
                            strS2 = "x" + strS2;
                            break;
                        }
                    }
                    k--;

                }
            }

            Console.WriteLine(intCnt);


        }
    }
}