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 C (GCC 5.4.1) (judge verdict: Accepted)

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>  
#include <stdlib.h> 
#include <string.h> 

//回文を判定する関数 内容の理解は頑張って
int IsPalindrome(char* cs) {
	int len = strlen(cs);
	for (int i = 0; i < len; i++)
	{
		if (cs[i] != cs[len - 1 - i]) return 0;
	}
	return 1;
}

int check() {
	char a[105], b[105], c[105];
	int len;
	scanf("%s", &a);
	len = strlen(a);

	// 文字列全体が回文になっているか調べる
	if (!IsPalindrome(a)) return 0;

	// 前半部について回文になっているか調べる
	for (int i = 0; i < len / 2; i++)
	{
		b[i] = a[i];
	}
	b[len / 2] = '\0';
	if (!IsPalindrome(b)) return 0;

	// 後半部について回文になっているか調べる(いらない)
	for (int i = 0; i < len / 2; i++)
	{
		b[i] = a[len/2 + 1 + i];
	}
	b[len / 2] = '\0';
	if (!IsPalindrome(c)) return 0;
	return 1;
}

int main()
{
	if (check()) printf("Yes\n");
	else printf("No\n");
	return 0;
}