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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.io.IOException;
import java.util.ArrayList;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) throws IOException {
//		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
//		String W	 = in.readLine();
//		int num = Integer.parseInt(W.split(" ")[0]);

		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		String S = sc.next();
		String T = sc.next();
		String ans ="";

		for(int i = 0;i<N; i++) {
			ans += S.charAt(i);
			ans += T.charAt(i);
		}




		System.out.println(ans);

	}

	//	//二分探索
	//k <= num となる最小の配列要素kのインデックスを返す
	static private int binarySearch(long num, long[] orderedArray){
		int lowerBorder = -1;
		int upperBorder = orderedArray.length;
		int mid;

		while(upperBorder - lowerBorder >1) {
			mid = (upperBorder + lowerBorder)/2;
			if(orderedArray[mid]<=num) {
				lowerBorder = mid;
			}else {
				upperBorder = mid;
			}
		}
		return lowerBorder;
	}

	//二分探索
	//k <= num となる最小のList要素kのインデックスを返す
	static private int binarySearch(long num, ArrayList<Long> orderedList){
		int lowerBorder = -1;
		int upperBorder = orderedList.size();
		int mid;

		while(upperBorder - lowerBorder >1) {
			mid = (upperBorder + lowerBorder)/2;
			if(orderedList.get(mid)<=num) {
				lowerBorder = mid;
			}else {
				upperBorder = mid;
			}
		}
		return lowerBorder;
	}

	//aとbの最小公倍数を求める
	public static int gcd(int a, int b) {
        return b == 0 ? a: gcd(b, a % b);
	}
	public static long gcd(long a, long b) {
        return b == 0 ? a: gcd(b, a % b);
	}



}