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 <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <queue>
#include <utility>
#include <map>
#define ll long long
using namespace std;

int main(void){
    string s;
    cin >> s;
    bool ans = true;
    ll count = (s.length()-1)/2-1;
    for(ll i = 0; i < (s.length()-1)/2; i++){
        if(s[i]!=s[count])
            ans = false;
        //cout << i << " " << count << endl;
        count--;
    }
    count = s.length()-1;
    for(ll i = (s.length()+3)/2-1; i < s.length(); i++){
        if(s[i]!=s[count])
            ans = false;
        //cout << i << " " << count << endl;
        count--;
    }
    count = s.length()-1;
    for(ll i = 0; i < s.length(); i++){
        if(s[i]!=s[count])
            ans = false;
        count--;
    }

    if(ans==true)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    return 0;
}