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 rep(i, n) for (int i = 0; i < n; i++)
#define INF 1001001001
#define LINF (1LL << 62)
using namespace std;
using ll = long long;
using P = pair<int, int>;

void solv() {
    string s;
    cin >> s;
    int len = s.size();
    rep(i, len / 2) {
        if (s[i] != s[len - i - 1]) {
            cout << "No" << endl;
            return;
        }
    }
    int center = len / 2;
    rep(i, center / 2) {
        if (s[i] != s[center-i-1]) {
            cout << "No" << endl;
            return;
        }
    }
    int diff = len - (len + 3) / 2;
    int start = (len + 3) / 2 -1;
    for (int i = 0; i < diff / 2; i++) {
        if (s[i + start] != s[len-i-1]) {
            cout << "No" << endl;
            return;
        }
    }
    cout << "Yes" << endl;
}

int main() {
    solv();
    return 0;
}