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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.util.Scanner

fun main(args: Array<String>){
    val scn = Scanner(System.`in`)
    val str = scn.nextLine()
    val reverse = str!!.reversed()
    val n = str.length

    if(n % 2 == 0 || n !in 3..99){
        print("No")
        return
    }

    val regex = Regex("[a-z]*")
    if(!regex.matches(str)){
        print("No")
        return
    }



    if(str!=reverse){
        print("No")
        return
    }

    val mae = str.substring(0..(((n-1)/2)-1))
    val reverseMae = mae.reversed()

    if(mae!=reverseMae){
        print("No")
        return
    }

    val ushiro = str.substring((((n+3)/2)-1)..(n-1))
    val reverseUshiro = ushiro.reversed()

    if(ushiro!=reverseUshiro){
        print("No")
        return
    }


    print("Yes")
}