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>
#define io std :: ios_base :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define M 1000000007
#define int long long
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define rall(v) (v).rbegin(),(v).rend()
#define nl "\n"
#define yy "YES\n"
#define nn "NO\n"
using namespace std;
bool ispalin(string s)
{
    int n=s.length();
    for(int i=0;i<n/2;i++)
    {
        if(s[i]!=s[n-i-1])
            return false;
    }
    return true;
}
signed main()
{
    io;
    string s;
    cin>>s;
    int i;
    int n=s.length();
    if(ispalin(s)){
    if(((n-1)/2)%2==0)
    {
        string tmp="",tmp1="";
        for(int i=0;i<n/2;i++)
            tmp+=s[i];
        for(int i=n/2+1;i<n;i++)
            tmp1+=s[i];
        if(ispalin(tmp1) && ispalin(tmp))
            cout<<"Yes";
        else
            cout<<"No";
    }
    else{
        string t1="",t2="";
        for(int i=0;i<n/2;i++)
            t1+=s[i];
        for(int i=n/2+1;i<n;i++)
            t2+=s[i];
        if(ispalin(t1) && ispalin(t2))
            cout<<"Yes";
        else
            cout<<"No";
    }
    }
    else
        cout<<"No";
}