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)

// Ver19.03
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define INF (1 << 29) * 2
#define LLINF 4545454545454545454
#define MOD 1000000007
#define ll long long
#define ull unsigned long long
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))
int upll(const void *a, const void *b) { return *(ll *)a < *(ll *)b ? -1 : *(ll *)a > *(ll *)b ? 1 : 0; }
int downll(const void *a, const void *b) { return *(ll *)a < *(ll *)b ? 1 : *(ll *)a > *(ll *)b ? -1 : 0; }
void sortup(ll *a, int n) { qsort(a, n, sizeof(ll), upll); }
void sortdown(ll *a, int n) { qsort(a, n, sizeof(ll), downll); }

int main()
{
  char s[100];
  scanf("%s", s);
  int len = strlen(s), partialLen = (len - 1) / 2, isYes = 1;

  for (int i = 0; i < (int)floor(len / 2); i++)
  {
    if (s[i] != s[len - 1 - i])
      isYes = 0;
  }
  for (int i = 0; i < (int)floor(partialLen / 2); i++)
  {
    if ((s[i] != s[partialLen - 1 - i]) || (s[i + partialLen + 1] != s[2 * partialLen - i]))
      isYes = 0;
  }
  puts(isYes ? "Yes" : "No");
}