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 n = readLine()!!.split(" ")
    val nlength = n[0].length
    var f = 0
    val b = (nlength - 1) / 4 - 1
    val a = (nlength - 1) / 2 - 1
    val l = n[0].toCharArray()
    for (i in 0..a) {
        var m = nlength - 1 - i
        if (l[i] != l[m]) {
            f = 1
        }
    }
    for (i in 0..b){
        var ll = (nlength - 1) / 2 - i - 1
        var lll = (nlength + 3) / 2 + i - 1
        var m = nlength - 1 - i
        if (l[i] != l[ll]){
            f = 1
        } else if (l[lll] != l[m]){
            f = 1
        }
    }
    if (f == 1){
        println("No")
    } else {
        println("Yes")
    }
}