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;
using ll = long long;

#define REP(i, n) for (int i = 0; i < (n); i++)

bool judge_kaibun(string S) {
    string S2(S);
    reverse(S2.begin(), S2.end());
    for (int i = 0; i < S.size(); i++) {
        if (S[i] != S2[i]) return false;
    }
    return true;
}

int main() {
    cin.tie(0);
    ios_base::sync_with_stdio(false);
    string s; cin >> s;
    int len = s.size();
    int b1 = (len - 1)/2;
    int b2 = (len + 3)/2;

    if (judge_kaibun(s.substr(0, b1)) && judge_kaibun(s.substr(b2-1))) cout << "Yes" << endl;
    else cout << "No" << endl;

    return 0;
}