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.
x = int(input())
y = input()
ct = {0:[0, 0, 0, 0], 1:[0, 0, 0, 0], 2:[1, 0, 0, 0], 3:[1, 1, 0, 0], 4:[3, 1, 0, 0], 5:[3, 1, 1, 0], 6:[4, 2, 1, 0], 7:[4, 2, 1, 1], 8:[7, 2, 1, 1], 9:[7, 4, 1, 1]}
real = {2:[1, 0, 0, 0], 3:[1, 1, 0, 0], 4:[3, 1, 0, 0], 5:[3, 1, 1, 0], 6:[4, 2, 1, 0], 7:[4, 2, 1, 1], 8:[7, 2, 1, 1], 9:[7, 4, 1, 1]}
ut = {2:[1, 0, 0, 0], 3:[1, 1, 0, 0], 5:[3, 1, 1, 0], 7:[4, 2, 1, 1]}
lx = [0, 0, 0, 0]
for i in y:
    i = int(i)
    bad = ct[i]
    lx[0] += bad[0]
    lx[1] += bad[1]
    lx[2] += bad[2]
    lx[3] += bad[3]

s = ''
if lx[-1] >= 1:
    copies = lx[-1]
    lx[0] -= 4*copies
    lx[1] -= 2*copies
    lx[2] -= copies
    lx[3] -= copies
    s += '7' * copies

if lx[-2] >= 1:
    copies = lx[-2]
    lx[0] -= 3*copies
    lx[1] -= copies
    lx[2] -= copies
    s += '5' * copies

if lx[-3] >= 1:
    copies = lx[-3]
    lx[0] -= copies
    lx[1] -= copies
    s += '3' * copies

if lx[-4] >= 1:
    copies = lx[-4]
    lx[0] -= copies
    s += '2' * copies


print(s)
