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)

fun main(args: Array<String>){
    val s = readLine()!!
    val front = s.substring(0, (s.length-1)/2)
    val end = s.substring((s.length+3)/2 - 1, s.length)
    val check1 = palindromeCheck(s)
    val check2 = palindromeCheck(front)
    val check3 = palindromeCheck(end)
    if(check1 && check2 && check3){
        println("Yes")
    }else{
        println("No")
    }
}

fun palindromeCheck(s: String): Boolean{
    val length = s.length
    val lastIndex = s.lastIndex
    if(length % 2 != 0) {
        for (i in 0..(length - 1) / 2) {
            if (s[i] != s[lastIndex - i]) {
                return false
            }
        }
    }else{
        for (i in 0..length / 2) {
            if (s[i] != s[lastIndex - i]) {
                return false
            }
        }
    }
    return true
}