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;
// ----------------------R00TDADA------------------------------
#define rep(i, a, b) for (__typeof((b)) i = (a); i < (b); i++)
#define nrep(i, a, b) for (__typeof((b)) i = (a); i > (b); i--)
#define all(a) (a).begin(), (a).end()
#define ff first
#define ss second
#define ppi pair<int, int>
#define pppi pair<ppi, int>
#define vi vector<int>
#define vii vector<ppi>
#define viii vector<pppi>
#define vs vector<string>
#define pb push_back
#define mp make_pair
#define sz(x) (int)x.size()
#define M_PI 3.14159265358979323846
#define rootdada ios_base::sync_with_stdio(false)
#define int long long
#define M 1000000007
#define MAX 100005
int A[MAX], tree[4 * MAX];
int power(int x, int p)
{
    if (p == 0)
        return 1;
    int y = power(x, p / 2);
    y *= y;
    y %= M;
    if (p % 2 == 0)
        return y;
    y *= x;
    y %= M;
    return y;
}
void build(int node, int start, int end)
{
    if (start == end)
        //Leaf node will have a single element
        tree[node] = start;
    else
    {
        int mid = (start + end) / 2;
        //Recurse on the left child
        build(2 * node, start, mid);
        //Recurse on the right child
        build(2 * node + 1, mid + 1, end);
        // Internal node will have the sum of both of its children
        tree[node] = tree[2 * node] + tree[2 * node + 1];
    }
}
void update(int node, int start, int end, int idx, int val)
{
    if (start == end)
    {
        //Leaf Node
        A[idx] = val;
        tree[node] += val;
    }
    else
    {
        int mid = (start - end) / 2 + end;
        if (start <= idx && idx <= mid)
        {
            // If idx is in the left child, recurse on the left child
            update(2 * node, start, mid, idx, val);
        }
        else
        {
            // if idx is in the right child, recurse on the right child
            update(2 * node + 1, mid + 1, end, idx, val);
        }
        // Internal node will have the sum of both of its children
        tree[node] = tree[2 * node] + tree[2 * node + 1];
    }
}
int query(int node, int start, int end, int l, int r)
{
    if (r < start || l > end)
    {
        // range represented by a node is completely outside the given range
        return 0;
    }
    if (l <= start && end <= r)
    {
        // range represented by a node is completely inside the given range
        return tree[node];
    }
    // range represented by a node is partially inside and partially outside the given range
    int mid = (start + end) / 2;
    int p1 = query(2 * node, start, mid, l, r);
    int p2 = query(2 * node + 1, mid + 1, end, l, r);
    return p1 + p2;
}
void Baap_ko_mat_sikha()
{
    int n;
    cin >> n;
    string s, t, ans;
    cin >> s >> t;
    rep(i, 0, n)
    {
        ans.pb(s[i]);
        ans.pb(t[i]);
    }
    cout << ans << endl;
}

signed main()
{

    rootdada;
    int t;
    t = 1;
    while (t--)
    {
        Baap_ko_mat_sikha();
    }
}