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()
N = len(s)
s_trim = ''.join([c for c in s if c != 'x'])
N_trim = len(s_trim)

def check_anagram():
    for i in range(N_trim):
        j = N_trim - 1 - i
        if s_trim[i] != s_trim[j]:
            return False
    return True

def solve():
    if not check_anagram():
        return -1

    # Center of anagram
    pr = N_trim // 2
    pl = pr
    if N_trim % 2 == 0:
        pl -= 1

    xl = 0
    tl = 0
    while xl < N:
        if s[xl] != 'x':
            assert s[xl] == s_trim[tl]
            if tl == pl:
                break
            tl += 1
        xl += 1

    if N_trim % 2 == 1:
        xr = xl
    else:
        xr = xl + 1
        while s[xr] == 'x':
            xr += 1

    cost = 0
    while True:
        assert s[xl] == s_trim[pl]
        assert s[xr] == s_trim[pr]
        assert s[xl] == s[xr]
        # assert s_trim[pl] == s_trim[pr]
        if s_trim[pl] != s_trim[pr]:
            while True: pass
        pl -= 1; pr += 1
        while True:
            xl -= 1; xr += 1
            if not (xl >= 0 and xr < N and s[xl] == 'x' and s[xr] == 'x'):
                break
        while xl >= 0 and s[xl] == 'x':
            cost += 1
            xl -= 1
        while xr < N and s[xr] == 'x':
            cost += 1
            xr += 1
        if pl < 0 and pr >= N_trim and xl < 0 and xr >= N:
            return cost

print(solve())