Drazil is playing a math game with Varda.

Let's define <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7020b94c48be2ee5b8eeca05f379c8911bb0e7c4.png" style="max-width: 100.0%;max-height: 100.0%;"/> for positive integer *x* as a product of factorials of its digits. For example, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/6784d861d9a724d0aeddefcb00b7d0c3442ae586.png" style="max-width: 100.0%;max-height: 100.0%;"/>.

First, they choose a decimal number *a* consisting of *n* digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number *x* satisfying following two conditions:

1. *x* doesn't contain neither digit 0 nor digit 1.

2. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7020b94c48be2ee5b8eeca05f379c8911bb0e7c4.png" style="max-width: 100.0%;max-height: 100.0%;"/> = <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/aa57e017a6c74d06b0a54bc47e3f6990e5af38c1.png" style="max-width: 100.0%;max-height: 100.0%;"/>.

Help friends find such number.
The first line contains an integer *n* (1 ≤ *n* ≤ 15) — the number of digits in *a*.

The second line contains *n* digits of *a*. There is at least one digit in *a* that is larger than 1. Number *a* may possibly contain leading zeroes.
def descompunere(nr):
    ls = []
    while nr % 2 == 0:
        ls.append(2)
        nr //= 2
    for i in range(3, nr + 1, 2):
        while nr % i == 0:
            nr //= i
            ls.append(i)
    return ls


def factorial(nr):
    l = []
    for i in range(2, nr + 1):
        l.extend(descompunere(i))
    return l


def refactorizare(l):
    def remake(nr):
        s = descompunere(nr)
        for i in s:
            l.pop(l.index(i))

    v = []
    while len(l) > 0:
        v.append(l[0])
        for i in range(l[0], 1, -1):
            try:
                l.pop(l.index(i))
            except ValueError:
                remake(i)
    return v


s = input()
s = input()

v = []
for i in s:
    v.extend(factorial(int(i)))
v.sort(reverse=True)
s = refactorizare(v)
for i in s:
    print(i, end="")
