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;
#define rep(i, n) for (int i=0; i<(int)(n); i++)
#define all(x) x.begin(), x.end()
#define mod 1000000007
typedef long long ll;

bool isKaibun(string str, int front, int back){
    rep(i, (back-front+1)/2){
        if(str[front+i] != str[back-i]) return false;
    }
    return true;
}

int main() {
    string str;
    cin >> str;
    int n = str.length();
    if(isKaibun(str, 0, n-1) && isKaibun(str, 0, ((n-1)/2)-1) && isKaibun(str, ((n+3)/2)-1, n-1)){
        cout << "Yes" << endl;
    }else{
        cout << "No" << endl;
    }
    return 0;
}