Problem:
Score : 200 points

Problem Statement

A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:

S is a palindrome.

Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.

The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.

Determine whether S is a strong palindrome.

Constraints

S consists of lowercase English letters.

The length of S is an odd number between 3 and 99 (inclusive).

Input

Input is given from Standard Input in the following format:

S

Output

If S is a strong palindrome, print Yes;
otherwise, print No.

Sample Input 1

akasaka

Sample Output 1

Yes

S is akasaka.

The string formed by the 1-st through the 3-rd characters is aka.

The string formed by the 5-th through the 7-th characters is aka.
All of these are palindromes, so S is a strong palindrome.

Sample Input 2

level

Sample Output 2

No

Sample Input 3

atcoder

Sample Output 3

No

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

#require "pry-byebug"
def is_kaibun(str)
  result = true
  ary = str.split("")
  if 2 < str.length
    loop do
      first = ary.shift
      last = ary.pop
      if first != last
        result = false
        break
      end
      if ary.length < 3
        break
      end
    end
  end
  result
end

S=gets.chomp
result = is_kaibun(S)
if result != true
  puts "No"
  exit
end

N = S.length
end_pos = (N - 1) / 2
s_sub = S.slice(0, end_pos)
result = is_kaibun(s_sub)
if result != true
  puts "No"
  exit
end

start_pos = (N - 1 + 3) / 2
#puts "S:#{S}, N:#{N}, start_pos:#{start_pos}, s_sub:#{s_sub}"
s_sub = S.slice(start_pos..N)
result = is_kaibun(s_sub)
if result != true
  puts "No"
  exit
end

puts "Yes"