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;
#define ff first
#define ss second
#define pb push_back
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1e9 + 7;
const ll inf = 9223372036854775807;
const ll mininf = -9223372036854775807;
const ll nax = 0;

ll n, cnt;
string s;
ll bl;
 
int main(){
	cin >> s;
	for (ll i = 0; i < s.length() / 2; i++){
		if (s[i] != s[s.length() - i - 1]){
			bl = 1;
		}
	}
	n = (s.length() - 1) / 2;
	// cout << n << '\n';
	for (ll i = 0; i < n / 2; i++){
		if(s[i] != s[n - i - 1]){
			bl = 2;
		}
	}
	n = (s.length() + 3) / 2;
	n--;	
	// cout << n << '\n';
	for (ll i = n; i <= s.length() - (s.length() - n); i++){
		if (s[i] != s[s.length() - cnt - 1]){
			bl = 3;
		}
		cnt++;
	}
	if (bl == 0){
		cout << "Yes" << '\n';
	}
	else{
		cout << "NO" << '\n';
	}
}