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>) {

    fun isPalindrome(s: String): Boolean {
        return if (s.length % 2 == 1) {
            val mae = s.substring(0 until s.length / 2)
            val usiro = s.substring(s.length / 2 + 1 until s.length)
            return mae == usiro.reversed()
        } else {
            val mae = s.substring(0 until s.length / 2)
            val usiro = s.substring(s.length / 2 until s.length)
            return mae == usiro.reversed()
        }
    }

    val str = readLine()!!

    if (!isPalindrome(str)) {
        println("No")
        return
    }

    if (!isPalindrome(str.substring(0 until (str.length - 1) / 2))) {
        println("No")
        return
    }

    if (!isPalindrome(str.substring(((str.length + 3) / 2) - 1 until str.length))) {
        println("No")
        return
    }

    println("Yes")
}