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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

//--- pB ---//
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main {
	public static void main (String[] args) throws java.lang.Exception {
		new Solution();
	}
}
class Solution {
	Reader reader;
	public Solution () {
		reader = new Reader();
		while (reader.hasNext()) {
			run_case();
		}
	}
	private void run_case() {
		String s = reader.ns();
		int N = s.length();
		String sa = s.substring(0, ((N-1)/2));
		String sb = s.substring( (N+3)/2 - 1 );

		if (isPalin(s) && isPalin(sa) && isPalin(sb)) {
			System.out.println("Yes");
		} else {
			System.out.println("No");
		}

		return;
	}
	private boolean isPalin(String s) {
		for (int i=0; i<s.length()/2; i++) {
			if (s.charAt(i) != s.charAt(s.length() - 1 - i)) return false;
		}
		return true;
	}
}

class Reader {
	BufferedReader br;
	StringTokenizer st;
	public Reader(){
		try {
			br = new BufferedReader(new InputStreamReader(System.in));
			st = new StringTokenizer(br.readLine());
		} catch (Exception e) {
			e.printStackTrace();
		}
	}
	public String next() {
		if (st.hasMoreTokens()) return st.nextToken();
		try {
			st = new StringTokenizer(br.readLine());
		} catch (Exception e) {
			e.printStackTrace();
		}
		return st.nextToken();
	}
	public boolean hasNext() {
		try {
			if (!st.hasMoreTokens() && !br.ready()) return false;
			else return true;
		} catch (Exception e) {
			e.printStackTrace();
		}
		return false;
	}
	public int ni() {return Integer.parseInt(this.next());}
	public String ns() {return this.next();}
}