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: Accepted)

package main

import (
	"fmt"
)

func main() {
	var s string
	fmt.Scanf("%s", &s)
	answer := isStrongParlindrome(s)
	if answer {
		fmt.Println("Yes")
	} else {
		fmt.Println("No")
	}
}

func isStrongParlindrome(s string) bool {
	return isParlindrome(s) && isConditionA(s) && isConditionB(s)
}

func isConditionA(s string) bool {
	start := 0
	end := (len(s) - 1) / 2
	return isParlindrome(s[start:end])
}

func isConditionB(s string) bool {
	start := (len(s)+3)/2 - 1
	end := len(s)
	return isParlindrome(s[start:end])
}

func isParlindrome(s string) bool {
	for i := 0; i < len(s)/2+1; i++ {
		if s[i] != s[len(s)-i-1] {
			return false
		}
	}
	return true
}