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 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <string.h>

#define N 100000

int main(void){
    char s[N], c;
    int len=0, p, q;
    int fhx[N], shx[N];
    int *cnt1, *cnt2;
    int ret=-1;

    while(isalnum(c = getchar()))
        s[len++] = c;

    p = 0;
    q = len-1;
    memset(fhx, -1, N);
    memset(shx, -1, N);
    cnt1 = fhx; cnt2 = shx;
    while(1){
        *cnt1=0; *cnt2=0;
        while(s[p] == 'x' && p<q){
            p++; (*cnt1)++;
        }
        while(s[q] == 'x' && p<q){
            q--; (*cnt2)++;
        }
        if(p >= q){
            cnt1=fhx; cnt2=shx;
            ret = 0;
            while(*cnt1 != -1 && *cnt2 != -1){
                ret += abs(*cnt1 - *cnt2);
                cnt1++; cnt2++;
            }
            break;
        }
        if(s[p] != s[q]) break;
        else{
            cnt1++; cnt2++;
            p++; q--;
        }
    }
    

    printf("%d\n", ret);

    return 0;
}