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 Python (3.4.3) (judge verdict: Accepted)

S = input()

cond1 = False #Sが回文
cond2 = False #前半分が回文
cond3 = False #後半分が回文

N = len(S)
end_full = int(((N+1)/2)-1)

if S[0:end_full]==S[::-1][0:end_full]:
  cond1 = True

half_N = len(S[0:end_full])
#print("half_N",half_N)

if half_N%2==0:
  half_end = int(half_N/2)
else:
  half_end = int(((half_N+1)/2)-1)
#print("half_end",half_end)

prevS = S[0:end_full]
#print("prevS1",prevS[0:half_end])
#print("prevS2",prevS[::-1][0:half_end])
if prevS[0:half_end]==prevS[::-1][0:half_end]:
  cond2 = True
  
suffS = S[::-1][0:end_full]
if suffS[0:half_end]==suffS[::-1][0:half_end]:
  cond3 = True
  
if cond1 and cond2 and cond3:
  print("Yes")
else:
  print("No")