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 judgeKaibun(s: String): Int {
    val end: Int = s.length
    for(i in 0..end/2) {
        if(s[i] != s[end - i - 1]) {
            println("No")
            return -1
        }
    }
    return 0
}
fun main(args: Array<String>) {
    val s: String = readLine()!!
    val n: Int = s.length
    if(judgeKaibun(s.substring(0, n)) !=  0) return
    if(judgeKaibun(s.substring(0, (n - 1) / 2)) !=  0) return
    if(judgeKaibun(s.substring((n + 3) / 2 - 1, n)) !=  0) return
    println("Yes")
}