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)

  long long A,B,C,D,E,F,    I,J,K,  M,N,O,P,Q,R,S,T,U,V,  X,Y,Z;
//long long A,B,C,D,E,F,  H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z;
double dA,dB,dC,dD,dE,dF,   dI,dJ,dK,  dM,dN,dO,dP,dQ,dR,dS,dT,dU,dV,  dX,dY,dZ;
#include <string>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <string>
#include <iomanip>
#include <cmath>

//typedef pair<long long,long long> P;
//typedef pair<long long,P> P1;
//typedef pair<P,P> P2;
#define pu push
#define pb push_back
#define mp make_pair
#define eps 1e-7
//#define INF 100000000000000000//19桁
#define INF 1e18+5//19桁
#define mod 1000000007
#define fi first
#define se second
#define rep(i,x) for(long long i=0;i<x;i++)
#define repn(i,x) for(long long i=1;i<=x;i++)
#define REP(i,n,x) for(long long i=n;i<x;i++)
#define REPN(i,n,x) for(long long i=n+1;i<x;i++)
#define repn(i,x) for(long long i=1;i<=x;i++)
#define rrep(i,x) for(long long i=x-1;i>=0;i--)
#define rrepn(i,x) for(long long i=x;i>1;i--)
#define SORT(x) sort(x.begin(),x.end())
#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())
#define All(x) (x).begin(),(x).end()
#define rAll(x) (x).rbegin(),(x).rend()

#define ALL(v)  (v).begin(),(v).end()
//#define MAX(a,b) if(a>b)
#define MAX(a,b) (((a)>(b))?(a):(b))
#define MIN(a,b) (((a)<(b))?(a):(b))


const string MINUSINF = "-";

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
void chmax(string &a, string b) {
    if (a == MINUSINF) a = b;
    else if (a.size() < b.size()) a = b;
    else if (a.size() == b.size()) {
        if (a < b) a = b;
    }
}


template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

#define pr printf
#define re return
#define CASET int ___T, case_n = 1; scanf("%d ", &___T); while (___T-- > 0)


int GCD(int a, int b) { return b ? GCD(b, a%b) : a; }

stack<long long> sta;   
queue<long long> que;
deque<long long> dque;

set <string> s_set;






long long sum,sum1,sum2,sum3,sum4;
long long flg,flg1,flg2,flg3,cnt,cnt1,cnt2,cnt3,cnt4;
long long wk,wk1,wk2,wk3,wk4;
long long ans;

double dsum,dsum1,dsum2,dsum3,dsum4;
double dcnt,dcnt1,dcnt2,dcnt3,dcnt4;
double dwk,dwk1,dwk2,dwk3,dwk4;
double dans;
double dmax,dmax1;
double dmin,dmin1;



bool f;

long long weight[110], value[110]; // 品物の個数は 100 個なので少し余裕持たせてサイズ 110 に
ll a[100010][3]; // a[i], b[i], c[i] をそれぞれまとめて a[i][0], a[i][1], a[i][2] にしてしまう
long long n,h[100005];
vector<long long> p, c;
int vis[200001];

vector<string> fi;
int Left[2100][2100], Right[2100][2100], Up[2100][2100], Down[2100][2100];

long long match[10] = {2,5,5,4,5,6,3,7,6};
vector<long long> box;



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

	long long max,max1;
	long long min,min1;
	

	sum=sum1=sum2=sum3=sum4=0;
	flg=flg1=flg2=flg3=cnt=cnt1=cnt2=cnt3=cnt4=0;
	max=max1=0;
	wk=wk1=wk2=wk3=wk4=0;
	ans=0;
	min=min1=INF;

	dsum=dsum1=dsum2=dsum3=dsum4=0;
	dcnt=dcnt1=dcnt2=dcnt3=dcnt4=0;
	dmax=dmax1=0;
	dwk=dwk1=dwk2=dwk3=dwk4=0;
	dans=0;
	dmin=dmin1=INF;
	
	

	//**************************こどふぉ用******************************************
	/*
	cin >> T;
	while(T--){
		cin >> N >> K;
		//std::vector<long long> v(N);	
		//vector<ll> v(N, 0);
		//set<long long> st;
		//map<std::string, int> freq;
		//map<int,int> M;
		//M.clear();
		
		//if(st.find(v[i])==st.end()){
		//	st.insert(v[i]);
		//}
		//else{
		//	st.erase(v[i]);
		//}
		
		//vector<pair<long long, long long> > p(M);
		//for(long long i=0; i<M; i++){
		//long long a, b;
		//cin >> a >> b;
		//p[i] = make_pair(a, b);
		//}
		//int ret=1<<20;
		
		for(int i=1;i<=N;i++){
			
		}
		cout << N << "\n";
	}
	*/
	
	//**************************ここからスタート******************************************
	std::cin >> N;
	//std::cin >> N >> M;
	//std::cin >> N >> M >> K;
	//std::cin >> N >> K;
	//std::cin >> A >> B;
	//std::cin >> A >> B >> C;
    
	//std::vector<long long> v(N);
	//std::vector<long long> w(N);
	//std::vector<long long> x(N);
	//std::vector<int> v(100, 0);
	//int a[26]={0};
	//vector<long long> flagment(N, 0);
	
	//v.push_back(i);
	//accumulate(vec.begin(), vec.end(), 0)
    //v.resize
    
	// 配列入力1
	
    string s,t,rr;
    cin >> s;
    cin >> t;
    
    rr=s+t;

    for(long long i=0; i<N; i++){
	     rr[2*i]=s[i];
         rr[2*i+1]=t[i];
    }
	std::cout << rr << std::endl;
	
    re 0;
}