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.
import sys
input = sys.stdin.readline

############ ---- Input Functions ---- ############
def inp(): # int
    return(int(input()))
def inlt(): # list
    return(list(map(int,input().split())))
def insr(): # string as char list
    s = input()
    return(list(s[:len(s) - 1]))
def instr(): # string
    return input()
def invr(): # spaced ints
    return(map(int,input().split()))


# 2!: 2. Replace 2
# 3!: 2, 3. Replace 3.
# 4!: 2^3, 3. Replace 2, 2, 3
# 5!: 2^3, 3, 5. Replace 5.
# 6!: 2^4, 3^2, 5. Replace 3, 5
# 7!: 2^4, 3^2, 5, 7. Replace 7
# 8!: 2^7, 3^2, 5, 7. Replace 2, 2, 2, 7
# 9!: 2^7, 3^4, 5, 7. Replace 2, 3, 3, 7
n = inp()
a = inp()
num_digs = [0] * 10 # num_digs[i] is the number of copies of i to include in final
for i in range(n):
    curr_dig = a % 10
    a //= 10
    if curr_dig == 2:
        num_digs[2] += 1
    elif curr_dig == 3:
        num_digs[3] += 1
    elif curr_dig == 4:
        num_digs[2] += 2
        num_digs[3] += 1
    elif curr_dig == 5:
        num_digs[5] += 1
    elif curr_dig == 6:
        num_digs[3] += 1
        num_digs[5] += 1
    elif curr_dig == 7:
        num_digs[7] += 1
    elif curr_dig == 8:
        num_digs[2] += 3
        num_digs[7] += 1
    elif curr_dig == 9:
        num_digs[2] += 1
        num_digs[3] += 2
        num_digs[7] += 1
ans = []
for i in range(9, 1, -1):
    ans += [str(i)] * num_digs[i]
print(''.join(ans))
