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(raw_input())
s = {1: 0, 2: 0, 3: 0, 5 : 0, 7 : 0}
num = raw_input()
for c in num:
    if c == '1':
        s[1] += 1
    elif c == '2':
        s[2] += 1
        s[1] += 1
    elif c == '3':
        s[3] += 1
        s[2] += 1
        s[1] += 1
    elif c == '4':
        s[3] += 1
        s[2] += 3
        s[1] += 1
    elif c == '5':
        s[5] += 1
        s[3] += 1
        s[2] += 3
        s[1] += 1
    elif c == '6':
        s[2] += 1
        s[3] += 1
        s[5] += 1
        s[3] += 1
        s[2] += 3
        s[1] += 1
    elif c == '7':
        s[7] += 1
        s[2] += 1
        s[3] += 1
        s[5] += 1
        s[3] += 1
        s[2] += 3
        s[1] += 1
    elif c == '8':
        s[2] += 3
        s[7] += 1
        s[2] += 1
        s[3] += 1
        s[5] += 1
        s[3] += 1
        s[2] += 3
        s[1] += 1
    elif c == '9':
        s[3] += 2
        s[2] += 3
        s[7] += 1
        s[2] += 1
        s[3] += 1
        s[5] += 1
        s[3] += 1
        s[2] += 3
        s[1] += 1
r = ''
if s[7]:
    r += '7' * s[7]
    s[2] -= 1 * s[7]
    s[3] -= 1 * s[7]
    s[5] -= 1 * s[7]
    s[3] -= 1 * s[7]
    s[2] -= 3 * s[7]
    s[7] -= s[7]
    s[1] -= 1
if s[5]:
    r += '5' * s[5]
    s[1] -= 1
    s[3] -= 1 * s[5]
    s[2] -= 3 * s[5]
    s[5] -= 1 * s[5]
if s[3]:
    r += '3' * s[3]
    s[1] -= 1
    s[2] -= 1 * s[3]
    s[3] -= 1 * s[3]
if s[2]:
    r += '2' * s[2]
    s[1] -= 1
    s[2] -= 1 * s[2]
print r
