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 9.2.1) (judge verdict: Accepted)

/*
 AUTHOR : ANSHU BABU RAJPUT
 
    JAI MAA SARASWATI
    OM NAMAH SHIVAY
    JAI SHRI KRISHNA
    JAI SHRI RAM
    BHARAT MATA KI JAI

    #PKMKB	
    #CKMKB
*/
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
using namespace std;

#define int  long long 
#define endl    ("\n")
#define ff       first
#define ss      second
#define pb   push_back
#define pf  push_front
#define ppb   pop_back
#define ppf  pop_front
#define mp   make_pair
#define mod 1000000007

#define L(x)    x.length()
#define B(x)     x.begin()
#define E(x)       x.end()
#define F(x)     x.front()
#define SZ(x)     x.size()
#define CLR(x)   x.clear()

#define all(x)                      x.begin(),x.end()
#define sortD(x) sort(x.begin(),x.end(),greater<int>)
#define rev(x)             reverse(x.begin(),x.end())

#define intV vector<int>

#define PI          (3.141592653589)
#define PMAX     priority_queue<int>
#define PMIN             priority_queue<int,vi,greater<int>>
#define setbits(x)                   __builtin_popcountll(x)
#define zrobits(x)                        __builtin_ctzll(x)

#define KING(i, n) for (int(i) = 0; (i) < n; (i)++)
#define MAHAKAL(i, n) for (int(i) = n-1; (i) >= 0; (i)--)

#define Anshu        int t; cin>>t; while(t--)
#define SITARAM(n,type) type* input=new type[n]

mt19937                          rng(chrono::steady_clock::now().time_since_epoch().count());
 
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;


void ViLiFY_KiNG()
{
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    freopen("error.txt","w",stderr);
#endif
}
bool isPalindrome(string str) {
    // bool temp = true;
    int i=0;
    int j=str.length()-1;

    while(i<=j) {
        if(str[i] != str[j]) {
            return false;
        }
        i++;
        j--;
    }
    return true;
}
int32_t main()
{
    ViLiFY_KiNG();
    string str;
    cin>>str;


    if(L(str)%2 == 0) {
        cout<<"No"<<endl;
      	return 0;
    }

    if(isPalindrome(str)) {
        if(isPalindrome(str.substr(0,(L(str)-1)/2))) {
            cout<<"Yes"<<endl;
            // cout<<str.substr(0,(L(str)-1)/2)<<endl;
        } else {
            cout<<"No"<<endl;
            // cout<<str.substr(0,(L(str)-1)/2)<<endl;
        }
    } else {
        cout<<"No"<<endl;
        // cout<<str.substr(0,(L(str)-1)/2)<<endl;
    }
    return 0;
}