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`.

Examples

Input

akasaka


Output

Yes


Input

level


Output

No


Input

atcoder


Output

No
import java.io.InputStream;
import java.io.PrintStream;
import java.util.Scanner;

public class Main {

  public static void main(String[] args) {
    resolve(System.in, System.out);
  }

  static void resolve(InputStream is, PrintStream pw) {
    Scanner sc = new Scanner(is);
    String s = sc.next();
    if (!isKaibun(s)) {
      pw.println("No");
      return;
    }
    if (!isKaibun(s.substring(0, (s.length() - 1) / 2))) {
      pw.println("No");
      return;
    }
    if (!isKaibun(s.substring((s.length() + 3 - 1) / 2))) {
      pw.println("No");
      return;
    }
    pw.println("Yes");
  }

  static boolean isKaibun(String s) {
    return s.equals(new StringBuilder((s)).reverse().toString());
  }
}
