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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
using P = pair<ll,ll>;
const ll mod = 1e9+7;
#define ALL(x) (x).begin(),(x).end()
#define pow(x,y) modpow(x,y)
#define REP(i,n) for(ll (i)=0;(i)<(n);(i)++)

template <class T = int>T in(){
    T x;
    cin >> x;
    return (x);
}

signed main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout<<fixed<<setprecision(10);
    string S = in<string>();
    bool ok = 1;
    int N = S.size();
    REP(i,N/2)if(S[i] != S[N-1-i])ok = 0;
    REP(i,(N/2)/2)if(S[i] != S[N/2-1-i])ok = 0;
    REP(i,(N/2)/2)if(S[N-1-i] != S[N/2+1-i])ok = 0;
    cout << (ok ? "Yes" : "No") << "\n";
}