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 <assert.h>
#include <limits.h>
#include <algorithm>
#include <bitset>
#include <cctype>
#include <cmath>
#include <complex>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <vector>
 
using ll = long long;
using P = std::pair<ll, ll>;
 
#define rep(i, a, b) for (ll(i) = (a); i < (b); i++)
#define all(i) i.begin(), i.end()
#define debug(i) std::cerr << "debug " << i << std::endl
 
template <typename T1, typename T2>
std::ostream& operator<<(std::ostream& os, std::pair<T1, T2> pa) {
  return os << pa.first << " " << pa.second;
}
 
template <typename T>
std::ostream& operator<<(std::ostream& os, std::vector<T> vec) {
  for (int i = 0; i < vec.size(); i++)os << vec[i] << (i + 1 == vec.size() ? "" : " ");
  return os;
}
 
template<typename T1,typename T2>
inline bool chmax(T1& a,T2 b){return a<b && (a=b,true);}
 
template<typename T1,typename T2>
inline bool chmin(T1& a,T2 b){return a>b && (a=b,true);}
 
ll pow_mod(ll a, ll b, ll mod=-1) {
  if ((a == 0)||(mod!=-1&&a%mod==0)) {
    return 0;
  }
 
  ll x = 1;
 
  while (b > 0) {
    if (b & 1) {
      x = (mod!=-1)?(x * a) % mod:x*a;
    }
    a = (mod!=-1)?(a * a) % mod:a*a;
    b >>= 1;
  }
  return x;
}
 
// const ll MOD = 998244353;
const ll MOD = 1e9 + 7;



int main() {
  std::cin.tie(0);
  std::ios::sync_with_stdio(false);

  ll n;
  std::cin>>n;

  std::string s;
  std::cin>>s;

  if(n%2==1){
    std::cout<<"No\n";
    return 0;
  }

  rep(i,0,n/2){
    if(s[i]!=s[i+n/2]){
      std::cout<<"No\n";
      return 0;
    }
  }

  std::cout<<"Yes\n";

  return 0;
}