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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using static System.Console;
using static System.Math;
 
public class Hello{
    public static void Main(){
        int k = rint();
        string[] read = stra();
        for(int i=0;i<k;i++){
            Write(read[0][i]);
            Write(read[1][i]);
        }
        WriteLine();
        
        
        
        
        
    }
    //各種読取
    static string rstr(){
        return ReadLine();
    }
    static int rint(){
        return int.Parse(ReadLine());
    }
    static long rlong(){
        return long.Parse(ReadLine());
    }
    static string[] stra(){
        return ReadLine().Split(' ');
    }
    static int[] inta(){
        string[] read_str_array = ReadLine().Split(' ');
        int[] return_int_array = new int[read_str_array.Length];
        for(int countup_i=0;countup_i<read_str_array.Length;countup_i++){
            return_int_array[countup_i] = int.Parse(read_str_array[countup_i]);
        }
        return return_int_array;
    }
    static int[,] inta2(int num_array,int in_array){
        int[,] int_array2 = new int[num_array,in_array];
        for(int i=0;i<num_array;i++){
            int[] temp_array = inta();
            for(int j=0;j<in_array;j++){
                int_array2[i,j] = temp_array[j];
            }
        }
        return int_array2;
    }
    // A/B小数点以下切上
    static int divideup(int divided,int divid){
        int temp_divide = divided/divid;
        if(divided % divid > 0){
            temp_divide++;
        }
        return temp_divide;
    }
    static int gcd(int big,int small){
        if(big < small){
            int temp = big;
            big = small;
            small = temp;
        }
        if(big % small == 0){
            return small;
        }else{
            big %= small;
            return gcd(small,big);
        }
    }
    static int lcm(int big,int small){
        long temp = big * small;
        int gcds = gcd(big,small);
        temp /= gcds;
        return (int)temp;
    }
}