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.
from math import sqrt,factorial
  

def primeFactors(n):
    r=[]
    while n % 2 == 0:
        r.append(2),
        n = n / 2

    for i in range(3,int(sqrt(n))+1,2):

        while n % i== 0:
            r.append(int(i)),
            n = n / i

    if n > 2:
        r.append(int(n))
          
    return r

n=int(input())
l=list(map(int,input()))
p=[]
for i in range(n):
    t=l[i]
    if t % 2 == 0:
        ll=primeFactors(t)
        if t >2:

            maks=ll[-1]
            if maks % 2 != 0:
                fak=factorial(maks)
                if fak == t:
                    p.append(maks)
                    p.append(t-1)
            else:
                y=len(ll)
                p+=[2]*y
                p.append(t-1)
        else:
            p.append(2)





    else:
        if t != 1:
            p.append(t)

p.sort(reverse=True)
print(*p,sep="")
