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 (1.3.71) (judge verdict: Wrong Answer)

fun main (args : Array<String>) {
    val s = readLine()!!

    var sList : MutableList<String> = mutableListOf()
    var fList : MutableList<String> = mutableListOf()
    var bList : MutableList<String> = mutableListOf()

    for (i in s) {
        sList.add(i.toString())
    }
    for (i in 0..( (sList.size - 1) / 2) - 1) {
        fList.add(sList[i])
    }
    for (i in ( (sList.size + 3) / 2 - 1)..sList.size - 1) {
        bList.add(sList[i])
    }

    for (i in 0..( (sList.size - 1) / 2) - 1) {
        if (fList[i] != bList[i]) {
            return println("No")
        }
    }

    for (i in 0..( (fList.size - 1) / 2) - 1) {
        if (fList[i] != fList[i + (fList.size / 2) + 1]) {
            return println("No")
        }
    }

    for (i in 0..( (bList.size - 1) / 2) - 1) {
        if (bList[i] != bList[i + (bList.size / 2) + 1]) {
            return println("No")
        }
    }

    return println("Yes")

}