largest_number

# Uses python3

import sys
import math

# didn't pass tests #8/11
def is_greater_or_equal5(n, current_max):
    # case n = 1000 any max will be greater except 1000!
    if len(n) == 4:
        if len(current_max) == 4:
            return True
        return False
    elif len(current_max) == 4:
        return True
    # otherwise if both numbers have same length a straight comparison works
    elif len(n) == len(current_max):
        return n >= current_max
    # other the possible pairs length are :
    # case A len(1,2)
    elif len(n) == 1 and len(current_max) == 2:
        if n[0] == current_max[0]:
            # we compare the second digit
            return n[0] >= current_max[1]
        return n[0] >= current_max[0]
    # case B len(1,3)
    elif len(n) == 1 and len(current_max) == 3:
        if n[0] == current_max[0]:
            # we compare the second digit
            if n[0] == current_max[1]:
                # we compare with the third digit
                return n[0] >= current_max[2]
            return n[0] >= current_max[1]
        return n[0] >= current_max[0]
    # case C len(2,1)
    elif len(n) == 2 and len(current_max) == 1:
        if n[0] == current_max[0]:
            # we compare the second digit
            return n[1] >= current_max[0]
        return n[0] >= current_max[0]
    # case D (2,3)
    elif len(n) == 2 and len(current_max) == 3:
        if n[0] == current_max[0]:
            # we compare with the second digit
            if n[1] == current_max[1]:
                # we compare with the last digit
                return n[1] >= current_max[2]
            return n[1] >= current_max[1]
        return n[0] >= current_max[0]
    # case E (3,1)
    elif len(n) == 3 and len(current_max) == 1:
        if n[0] == current_max[0]:
            # we compare the second digit
            if n[1] == current_max[0]:
                # we compare the third digit
                if n[2] == current_max[0]:
                    return True
                return n[2] >= current_max[0]
            return n[1] >= current_max[0]
        return n[0] >= current_max[0]
    # case F (3,2)
    elif len(n) == 3 and len(current_max) == 2:
        if n[0] == current_max[0]:
            # we compare the second digit
            if n[1] == current_max[1]:
                # we compare the third digit.
                if n[2] == current_max[1]:
                    return True
                return n[2] >= int(current_max[1])
            return n[1] >= current_max[1]
        return n[0] >= current_max[0]


# did pass all the tests thanks Mo Xin's hint
def compare_strings(s1, s2):
    first = s1+s2
    last = s2+s1
    return first >= last


def largest_number(a):
    # straight forward implementation of pseudo code given in assignment
    res = ""
    while a:
        max_digit = "0"
        for x in a:
            if compare_strings(x, max_digit):
                max_digit = x
        res += max_digit
        a.remove(max_digit)
    return res


if __name__ == '__main__':
    user_input = sys.stdin.read()
    data = user_input.split()
    digits = data[1:]
    print(largest_number(digits))