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 (GCC 5.4.1) (judge verdict: Accepted)

/*
   THE DARK NIGHT RISES
I LOVE YOU 72 AND WILL FOREVER DO SO
Patience, Persistence, Perseverance, Passion, Purpose
*/
#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ld;
const int N = 2e5 + 20 ;
#define F first
#define S second
# define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
#define pb push_back
#define mk make_pair
#define fast ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define rep(i,a,b) for(ll i=a;i<b;i++)
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<vi> vvi;
typedef vector<vii> vvii;
#define sz(x)       (int)x.size()
const long long int mod=1e9 + 7;
const long double EPS=1e-12;
#define mem( a, val ) memset(a, val, sizeof( a ) )
#define deci( x ) cout<<fixed<<setprecision( x );
#define bitcount( x ) __builtin_popcountll( x )
 
const int dx[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
const int dy[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
 
bool comp(pair<ll,ll> v1 ,pair<ll,ll> v2 )
{
 
}
bool vis[10] ;
 
int solve()
{
 	ll n ;
 	cin >> n ;
 	string s ,t ;
 	cin >> s >> t ;
 	int i = 0 , j = 0 ;
 	while(i < n && j < n)
 		cout << s[i++] << t[j++];
}
int main()
{
	fast;
	ll t;
	t=1;
	//cin>>t;
	while(t--)
	{
		solve();
	}
 
	return 0;
}
// Patience, Persistence, Perseverance, Passion, Purpose
// I LOVE YOU 72 AND WILL FOREVER DO SO.