Problem:
Score : 200 points

Problem Statement

Given are strings s and t of length N each, both consisting of lowercase English letters.

Let us form a new string by alternating the characters of S and the characters of T, as follows: the first character of S, the first character of T, the second character of S, the second character of T, ..., the N-th character of S, the N-th character of T. Print this new string.

Constraints

1 \leq N \leq 100

|S| = |T| = N

S and T are strings consisting of lowercase English letters.

Input

Input is given from Standard Input in the following format:

N
S T

Output

Print the string formed.

Sample Input 1

2
ip cc

Sample Output 1

icpc

Sample Input 2

8
hmhmnknk uuuuuuuu

Sample Output 2

humuhumunukunuku

Sample Input 3

5
aaaaa aaaaa

Sample Output 3

aaaaaaaaaa

Solution in C++14 (Clang 3.8.0) (judge verdict: Accepted)

#define _USE_MATH_DEFINES
#include <iostream>
#include <string>
#include <list>
#include <vector>
#include <queue>
#include <algorithm>
#include <climits>
#include <cstring>
#include <cmath>
#include <stack>
#include <iomanip>
#include <tuple>
#include <functional>
#include <cfloat>
#include <map>
#include <set>
#include <array>
#include <stdio.h>
#include <string.h>
using ll = long long;
using ull = unsigned long long;
using namespace std;
#define int long long
#define CONTAINS(v,n) (find((v).begin(), (v).end(), (n)) != (v).end())
#define SORT(v) sort((v).begin(), (v).end())
#define RSORT(v) sort((v).rbegin(), (v).rend())
#define ARY_SORT(a, size) sort((a), (a)+(size))
#define REMOVE(v,a) (v.erase(remove((v).begin(), (v).end(), (a)), (v).end()))
#define REVERSE(v) (reverse((v).begin(), (v).end()))
#define ARY_REVERSE(v,a) (reverse((v), (v)+(a)))
#define REP(i, n) for (int (i)=0; (i) < (n); (i)++)
#define CONTAINS_MAP(m, a) (m).find((a)) != m.end()
void YesNo(bool b) { cout << (b ? "Yes" : "No"); }
void Yes() { cout << "Yes"; }
void No() { cout << "No"; }
//max_element

signed main()
{
	int N;
	string S, T;
	cin >> N;
	cin >> S >> T;
	string ans;
	for (int i = 0; i < N; i++)
	{
		ans += S[i];
		ans += T[i];
	}
	cout << ans << endl;
}