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 count(digit):
    global l
    for i in range(0,digit-1):
        for d in div[i]:
            l[d-2]+=1


div = [[2],[3],[2,2],[5],[2,3],[7],[2,2,2],[3,3]] #descompunerea in factori primi a cifrelor
n = int(input())
number = input()
l = [0]*8
for i in number:
    if int(i) > 1:
        count(int(i))
res = ""

for i in range(7,-1,-1):
    for _ in range(0,l[i]):
        for k in range(0,i):
            #print(i+2,k+2)
            if len(div[k])>1:
                for j in div[k]:
                    l[j-2] -=1
                    #print(l)
            else:
                l[k]-=1

#print(l)
for i in range(7,-1,-1):
    res += (str(i+2)*l[i])
print(res)

