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 Rust (1.15.1) (judge verdict: Accepted)

macro_rules! input {
  (source = $s:expr, $($r:tt)*) => {
      let mut iter = $s.split_whitespace();
      let mut next = || { iter.next().unwrap() };
      input_inner!{next, $($r)*}
  };
  ($($r:tt)*) => {
      let stdin = std::io::stdin();
      let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
      let mut next = move || -> String{
          bytes
              .by_ref()
              .map(|r|r.unwrap() as char)
              .skip_while(|c|c.is_whitespace())
              .take_while(|c|!c.is_whitespace())
              .collect()
      };
      input_inner!{next, $($r)*}
  };
}

macro_rules! input_inner {
  ($next:expr) => {};
  ($next:expr, ) => {};

  ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
      let $var = read_value!($next, $t);
      input_inner!{$next $($r)*}
  };

  ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
    let mut $var = read_value!($next, $t);
    input_inner!{$next $($r)*}
  };
}

macro_rules! read_value {
  ($next:expr, ( $($t:tt),* )) => {
      ( $(read_value!($next, $t)),* )
  };

  ($next:expr, [ $t:tt ; $len:expr ]) => {
      (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
  };

  ($next:expr, chars) => {
      read_value!($next, String).chars().collect::<Vec<char>>()
  };

  ($next:expr, usize1) => {
      read_value!($next, usize) - 1
  };

  ($next:expr, $t:ty) => {
      $next().parse::<$t>().expect("Parse error")
  };
}

pub trait LexicalPermutation {
    /// Return `true` if the slice was permuted, `false` if it is already
    /// at the last ordered permutation.
    fn next_permutation(&mut self) -> bool;
    /// Return `true` if the slice was permuted, `false` if it is already
    /// at the first ordered permutation.
    fn prev_permutation(&mut self) -> bool;
}

impl<T> LexicalPermutation for [T]
where
    T: Ord,
{
    /// Original author in Rust: Thomas Backman <serenity@exscape.org>
    fn next_permutation(&mut self) -> bool {
        // These cases only have 1 permutation each, so we can't do anything.
        if self.len() < 2 {
            return false;
        }

        // Step 1: Identify the longest, rightmost weakly decreasing part of the vector
        let mut i = self.len() - 1;
        while i > 0 && self[i - 1] >= self[i] {
            i -= 1;
        }

        // If that is the entire vector, this is the last-ordered permutation.
        if i == 0 {
            return false;
        }

        // Step 2: Find the rightmost element larger than the pivot (i-1)
        let mut j = self.len() - 1;
        while j >= i && self[j] <= self[i - 1] {
            j -= 1;
        }

        // Step 3: Swap that element with the pivot
        self.swap(j, i - 1);

        // Step 4: Reverse the (previously) weakly decreasing part
        self[i..].reverse();

        true
    }

    fn prev_permutation(&mut self) -> bool {
        // These cases only have 1 permutation each, so we can't do anything.
        if self.len() < 2 {
            return false;
        }

        // Step 1: Identify the longest, rightmost weakly increasing part of the vector
        let mut i = self.len() - 1;
        while i > 0 && self[i - 1] <= self[i] {
            i -= 1;
        }

        // If that is the entire vector, this is the first-ordered permutation.
        if i == 0 {
            return false;
        }

        // Step 2: Reverse the weakly increasing part
        self[i..].reverse();

        // Step 3: Find the rightmost element equal to or bigger than the pivot (i-1)
        let mut j = self.len() - 1;
        while j >= i && self[j - 1] < self[i - 1] {
            j -= 1;
        }

        // Step 4: Swap that element with the pivot
        self.swap(i - 1, j);

        true
    }
}

#[cfg(not(debug_assertions))]
macro_rules! mydbg {
    ($($arg:expr),*) => {};
}

macro_rules! echo {
    ($($a:expr),*) => {
          $(println!("{}",$a))*
      }
}

use std::cmp::Ordering::*;
#[allow(unused_imports)]
use std::cmp::{max, min};
use std::collections::HashMap;
use std::io::{stdout, BufWriter, Write};
use std::mem::swap;

fn main() {
    input! {
        n: usize,
        s: chars,
        t: chars
    }

    for i in 0..n {
        print!("{}", s[i]);
        print!("{}", t[i]);
    }
}