Drazil is playing a math game with Varda.

Let's define <image> for positive integer x as a product of factorials of its digits. For example, <image>.

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. <image> = <image>.

Help friends find such number.

Input

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.

Output

Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation.

Examples

Input

4
1234


Output

33222


Input

3
555


Output

555

Note

In the first case, <image>
n=int(input())
a1=input()
a=int(a1)
sa=str(a)
numberl=[]
ans=""
for i in range(len(str(a))):
    if sa[i] == '0' or sa[i] == '1':
        continue
    elif sa[i] == '2':
        numberl.append(2)
    elif sa[i] == '3':
        numberl.append(3)
        numberl.append(2)
    elif sa[i] == '4':
        numberl.append(3)
        numberl.append(2)
        numberl.append(2)
        numberl.append(2)
    elif sa[i] == '5':
        numberl.append(5)
    elif sa[i] == '6':
        numberl.append(3)
        numberl.append(2)
        numberl.append(5)
    elif sa[i] == '7':
        numberl.append(7)
    elif sa[i] == '8':
        numberl.append(7)
        numberl.append(2)
        numberl.append(2)
        numberl.append(2)
    elif sa[i] == '9':
        numberl.append(3)
        numberl.append(3)
        numberl.append(2)
        numberl.append(2)
        numberl.append(2)
        numberl.append(7)
c2=numberl.count(2)
c3=numberl.count(3)
c5=numberl.count(5)
c7=numberl.count(7)
ans+="7"*c7
ans+="5"*c5
ans+="3"*c3
ans+="2"*(c2-c3)
ans1=list(ans)
ans2=sorted(ans1,reverse= True)
#print(ans2)
res = ""
for i in ans2:
    res+=i
print(int(res))