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 Ruby (2.3.3) (judge verdict: Runtime Error)

s = gets.chomp
s_rev = s.reverse
n = s.length
if n == 1
  puts 0
  exit
end
a = []
b = []
n.times do |i|
  if s[i] != "x"
    a << s[i]
    b << i
  end
end
a_rev = a.reverse
m = a.length

if m.odd?
  if a[0..((m-1)/2 - 1)] == a_rev[0..((m-1)/2 - 1)]
    left = s[0...b[(m-1)/2 - 1]]
    right = s[(b[(m-1)/2 - 1] + 1)..(n-1)].reverse
    if left.length == right.length
      ans = 0
      (n / 2).times do |i|
        ans += 1 if left[i] != right[i]
      end
      puts ans
    else
      puts (left.length - right.length).abs
    end
  else
    puts -1
  end
else
  if a[0...(m / 2)] == a_rev[0..(m / 2 - 1)]
    left = s[0..b[m/2 - 1]]
    right = s[b[m/2]..(n-1)].reverse
    if left.length == right.length
      ans = 0
      (n / 2).times do |i|
        ans += 1 if left[i] != right[i]
      end
      puts ans
    else
      puts (left.length - right.length).abs
    end
  else
    puts -1
  end
end