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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
/* typedef */
typedef long long ll;
/* constant */
const int INF = 1 << 30;
const ll LINF = 1LL << 50;
const int NIL = -1;
const int MAX = 10000;
const int mod = 1000000007;
const double pi = 3.141592653589;
/* global variables */
string S;
/* function */
bool isPalimdrome(int R, int L) {
    int r = R, l = L;
    while (r <= l) {
        if (S[r] != S[l]) return false;
        r++;
        l--;
    }
    return true;
}
/* main */
int main(){
    cin >> S;
    int len = S.size();
    if (isPalimdrome(0, (len - 1) / 2 - 1)
        && isPalimdrome(0, len - 1)
        && isPalimdrome((len + 3) / 2 - 1, len - 1))
        cout << "Yes" << '\n';
    else cout << "No" << '\n';
}