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 PyPy3 (2.4.0) (judge verdict: Accepted)

#coding:utf-8
import sys,os
from collections import defaultdict, deque
from fractions import gcd
from math import ceil, floor
sys.setrecursionlimit(10**6)
write = sys.stdout.write
dbg = (lambda *something: print(*something)) if 'TERM_PROGRAM' in os.environ else lambda *x: 0
def main(given=sys.stdin.readline):
    input = lambda: given().rstrip()
    LMIIS = lambda: list(map(int,input().split()))
    II = lambda: int(input())
    XLMIIS = lambda x: [LMIIS() for _ in range(x)]
    YN = lambda c : print('Yes') if c else print('No')
    MOD = 10**9+7

    S = input()
    N = len(S)
    def isP(S):
        return S == S[::-1]
    YN(isP(S) and isP(S[:(N-1)//2]) and isP(S[((N+3)//2-1):]))
    

if __name__ == '__main__':
    main()