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 bisect import *
m=int(input())
n=int(input())
prime=[2,3,5,7]
a=[0 for i in range(10)]
tmp=n
while tmp>0:
    r=tmp%10
    a[r]+=1
    tmp//=10
a[1]=0
digits=[]
ver=True
factors=[[0],[1],[2],[3],[2,2],[5],[3,2],[7],[2,2,2],[3,3]]
x={2:0,3:0}
for i in range(9,1,-1):
    if a[i]==0:
        continue
    if i==2 or i==3:
        for j in range(a[i]):
            if i==2:
                x[i]+=1
            else:
                x[i]+=1
                x[i-1]+=1
    elif i in prime:
        cnt=a[i]
        for j in range(cnt):
            digits.append(i)
    else:
        pos=bisect_left(prime,i)-1
        for j in range(a[i]):
            digits.append(prime[pos])
        for l in range(a[i]):
            for j in range(i,prime[pos],-1):
                for k in factors[j]:
                    x[k]+=1
for i in range(x[3]):
    digits.append(3)
x[2]-=x[3]
for i in range(x[2]):
    digits.append(2)
if ver:
    digits.sort(reverse=True)
    for i in digits:
        print(i,end="")
else:
    print(n)