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 Python (3.4.3) (judge verdict: Wrong Answer)

S = list(input())
invS = []
for s in reversed(S):
    invS.append(s)
cnt = 0
loop = 0
i,j = 0,0
while i < len(S)//2+1 and j < len(S)//2+1:
    if S[i] == invS[j]:
        i += 1
        j += 1
    elif S[i] == "x":
        cnt += 1
        i += 1
    elif invS[j] == "x":
        cnt += 1
        j += 1
    loop += 1
    if i == len(S)//2+1 and j == len(S)//2+1 and loop > len(S):
        cnt = -1
        break
print(cnt)