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 (2.7.6) (judge verdict: Accepted)

s = raw_input()

def precheck():
    res = ""
    for c in s:
        if c != "x":
            res += c
    if not res:
        return 1
    if res != "".join(reversed(res)):
        return 0
    return res



def calc(ns):
    lf = ""
    rf = ""
    
    vtype = len(ns) % 2

    pos = 0

    cnt = 0
    while cnt < len(ns)/2:
        lf += s[pos]
        if s[pos] == "x":
            pass
        else:
            cnt+=1
        pos+=1
    
    if vtype:
        while s[pos] == "x":
            lf += s[pos]
            pos+=1
        pos+=1
    else:
         while s[pos] == "x":
            pos+=1

   

    rf = "".join(reversed(s[pos:]))


    lx = 0
    rx = 0
    res = 0
    while 1:

        if lx >= len(lf):
            res += len(rf)-rx
            break
        if rx >= len(rf):
            res += len(lf)-lx
            break

        rrr = rf[rx] == "x"
        lll = lf[lx] == "x"

        if rrr and lll:
            rx+=1
            lx+=1
        elif rrr and not lll:
            rx += 1
            res += 1
        elif not rrr and lll:
            lx += 1
            res += 1
        else:
            rx += 1
            lx += 1
    return res

pc = precheck()
if pc == 1:
    print 0
elif not pc:
    print -1
else:
    print calc(pc)