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: Runtime Error)

s = input()
count = 0
if len(set(s)) == 1:
    print(0)
    exit()
left = 0
right = len(s) - 1

ans = 0
while True:
    if (s[left] == 'x' and s[right] == 'x') or s[left] == s[right]:
        left += 1
        right -= 1
    elif s[left] == 'x' and s[right] != 'x':
        ans += 1
        left += 1
    elif s[left] != 'x' and s[right] == 'x':
        ans += 1
        right -= 1
    elif s[left] != 'x' and s[right] != 'x' and s[left] != s[right]:
        print(-1)
        exit()
    if left == right:
        break
print(ans)