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 str = readLine()?.orEmpty()
    var result = "No"
    val sLength = str!!.length

    val strLeft = str.substring(0,((sLength - 1) / 2))
    val strRight = str.substring(((sLength + 1) / 2),sLength)

    if(judgeGram(str) && judgeGram(strLeft) && judgeGram(strRight)){
        result = "Yes"
    }

    
    println(result)
}

fun judgeGram(s:String):Boolean{
    var result = false
    val sLength = s.length

    var sLeft = ""
    var sRight = ""

    if(sLength % 2 == 0){
        sLeft = s.substring(0,sLength  / 2)
        sRight = s.substring(sLength / 2,sLength)

    }else{
        sLeft = s.substring(0,((sLength - 1) / 2))
        sRight = s.substring(((sLength + 1) / 2),sLength)
    }

    val sRR = sRight.reversed()
    if(sLeft == sRR){
        result = true
    }
    return result
}