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 Go (1.6) (judge verdict: Wrong Answer)

package main
 
import (
	"fmt"
)
 
func main() {
	var s string
	fmt.Scan(&s)
	s1 := s[:(len(s)-1)/2]
	s2 := s[(len(s)+3-1)/2:]
	f := equal(s1, s2)
	if f {
		fmt.Println("Yes")
	} else {
		fmt.Println("No")
	}
}
 
func equal(s1, s2 string) bool {
	if len(s1) == 1 {
		return true
	} else if len(s1) == 2 {
		if s1[0] == s1[1] && s2[0] == s2[1] {
			return true
		} else {
			return false
		}
	} else {
		for i := 0; i < len(s1); i++ {
			if s1[i] != s2[i] {
				return false
			}
		}
        for i, j :=0,len(s2)-1;; i, j = i+1, j-1 {
          if len(s1)%2 == 1 && i == j {
            break
          } else if len(s1)%2 == 0 && i > j {
            break
          }
          if s1[i] != s1[j] || s2[i] != s2[j] {
            return false
          }
      }
	}
 
	return true
}