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)

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef long double LD;

#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define X first
#define Y second
#define pb push_back
#define make_unique(x) sort(all(x)), x.resize(unique(all(x)) - x.begin())
#define int long long

const LL mod = 1e9 + 7;


template<class T>
istream& operator >> (istream& in, vector<T>& v){ for (auto &x : v) { in >> x; } return in; }

template<class T, class U>
istream& operator >> (istream& in, pair<T, U> & v){ in >> v.X >> v.Y;return in; }

template<class T, class U>
ostream& operator << (ostream& out, pair<T, U> & v){ out << v.X << " " << v.Y;return out; }

void chkmax(int &a, int b) {
    a = max(a, b);
    return;
}

void chkmin(int &a, int b) {
    a = min(a, b);
    return;
}

LL ppow (LL x, LL s) {
    if (!s) return 1;
    if (!(s - 1)) return x % mod;
    if (s % 2) return (x * ppow (x, s - 1)) % mod;
    LL b = ppow (x, s / 2);
    return (b * b) % mod;
}

main(){
    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    //freopen("ILUXA.in", "r", stdin);

    int n;
    string s, t;
    cin >> n >> s >> t;
    for (int i = 0;i < n;i++) {
        cout << s[i] << t[i];
    }
    return 0;
}