Problem:
Score : 200 points

Problem Statement

Given are a positive integer N and a string S of length N consisting of lowercase English letters.

Determine whether the string is a concatenation of two copies of some string.
That is, determine whether there is a string T such that S = T + T.

Constraints

1 \leq N \leq 100

S consists of lowercase English letters.

|S| = N

Input

Input is given from Standard Input in the following format:

N
S

Output

If S is a concatenation of two copies of some string, print Yes; otherwise, print No.

Sample Input 1

6
abcabc

Sample Output 1

Yes

Let T = abc, and S = T + T.

Sample Input 2

6
abcadc

Sample Output 2

No

Sample Input 3

1
z

Sample Output 3

No

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h> 

using namespace std;

#define YES cout << "YES" << endl;
#define NO  cout << "NO"  << endl;
#define Yes cout << "Yes" << endl;
#define No  cout << "No"  << endl;
 
#define INF INT_MAX
#define MOD 1000000007
#define PI  acos(-1)

using ll  = long long;
using ull = unsigned long long;
using Pii = pair<int, int>;
using Pll = pair<ll, ll>;

#define rep(i, n) for (ll i = 0; i < (n); i++)
#define rrep(i, n) for (ll i = (n)-1; i >= 0; i--)

int main(int argc, char *argv[])
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    ll N;
    cin >> N;

    string S;
    cin >> S;

    if (N%2 == 0) {
        string tmp1 = S.substr(0, N/2);
        string tmp2 = S.substr(N/2, N/2);
        if (tmp1 == tmp2) {
            Yes;
        } else {
            No;
        }
    } else {
        No;
    }

    return 0;
}