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;
using vl = vector<ll>;
using vvl = vector<vl>;
using vb = vector<bool>;
using P = pair<ll, ll>;
template <typename T> using pq = priority_queue<T>;
template <typename T> using minpq = priority_queue<T, vector<T>, greater<T>>;
template <typename T, typename K> using ump = unordered_map<T, K>;
const ll dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
const ll mod = 1000000007;
const ll inf = ll(1e9);
const ll e5 = ll(1e5);
const ll ll_inf = ll(1e9) * ll(1e9);

#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define repr(i, n) for (ll i = ll(n - 1); i >= 0; i--)
#define each(i, mp) for (auto &i : mp)
#define eb emplace_back
#define F first
#define S second
#define all(obj) (obj).begin(), (obj).end()

template <class T> ostream &operator<<(ostream &out, const vector<T> &list) {
    ll n = list.size();
    rep(i, n) out << list[i] << ' ';
    return out;
}

template <class T> istream &operator>>(istream &in, vector<T> &list) {
    ll n = list.size();
    rep(i, n) in >> list[i];
    return in;
}

template <class T> ostream &operator<<(ostream &out, const vector<vector<T>> &list) {
    ll n = list.size();
    rep(i, n) out << list[i] << '\n';
    return out;
}

/* ------------- ANSWER ------------- */
/* ---------------------------------- */

void solve() {
    string s;
    cin >> s;
    string t1, t2;
    ll n = s.size();
    rep(i, (n - 1) / 2) {
        t1 += s[i];
        t2 += s[i];
    }
    reverse(all(t2));
    if (t1 == t2) {
        string t1, t2;
        for (ll i = (n + 3) / 2 - 1; i < n; i++) {
            t1 += s[i];
            t2 += s[i];
        }
        reverse(all(t2));
        if (t1 == t2) {
            cout << "Yes" << '\n';
            return;
        }
    }
    cout << "No" << '\n';
}

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