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>
a = int(raw_input())
n = raw_input()
string = str(n)
b = []
for i in string:
    b.append(int(i))
c = []    

t = 0
while t <= a - 1:
    if b[t] != 1 and b[t] != 0 :
        c.append(b[t])
    t = t + 1          
r = 0
while r <= len(c) - 1:
    if c[r] == 2:
        c[r] == 2
    if c[r] % 2 == 1 and c[r] != 9:
        c[r] = c[r] 
    if c[r] == 9:
        c.remove(9)
        c.append(7)
        c.append(3)
        c.append(3)
        c.append(2)
    if c[r] % 2 == 0 and c[r] != 2:
        if c[r] == 4:
            c.remove(4)
            c.append(3)
            c.append(2)
            c.append(2)
        if c[r] == 6:
            c.remove(6)
            c.append(5)
            c.append(3)
        if c[r] == 8:
            c.remove(8)
            c.append(7)
            c.append(2)
            c.append(2)
            c.append(2)       
    r = r + 1
    a = 0
    b = 0
    if r == len(c):
        for i in c:
            if i == 2:
                a = 0
            if i == 3:
                a = 0
            if i == 4:
                b = 1
            if i == 5:
                a = 0
            if i == 6:
                b = 1
            if i == 7:
                a = 0
            if i == 8:
                b = 1
            if i == 9:
                b = 1
        if a+b == 0:
            r = len(c)
        else:
            r = 0
                
            
            
l = 0
p = 0
X = len(c) - 1
d = []
k = 0
while k <= X:
    while p <= len(c) - 1:
        if c[p] > l:
            l = c[p]
        p = p + 1
    d.append(l)
    c.remove(l)
    k = k + 1
    p = 0
    l = 0  
x = 0
string = ''
while x <= len(d) - 1:
    string = string + str(d[x])
    x = x + 1
print string    
