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 find(n,k): # finding k prime factor in n!
    ans=0
    i=1
    while n//(k**i) > 0 :
        ans+=(n//(k**i))
        i+=1
    return(ans)


def chebishev(n): # finding all prime factors in n!
    ans={2:0 , 3:0 , 5:0 , 7:0}
    for i in ans:
        ans[i]+=find(n,i)
    return(ans)


if __name__ == '__main__':

    n=int(input())
    s=input()

    q={2:0 , 3:0 , 5:0 , 7:0}

    for i in s:
        i=int(i)
        if i==0 or i==1 :
            continue
        y=chebishev(i)
        for i in y:
            q[i]+=y[i]

    t=sorted(q)
    t.reverse()

    answer=[]

    for i in t:
        while q[i]>0 :
            answer.append(str(i))
            o=chebishev(i)
            for j in o:
                q[j]-=o[j]
    
    answer.sort()
    answer.reverse()
    print(''.join(answer))



