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>
import sys
import string
import math
from collections import defaultdict
from functools import lru_cache
from collections import Counter
from fractions import Fraction

def mi(s):
    return map(int, s.strip().split())

def lmi(s):
    return list(mi(s))

def tmi(s):
    return tuple(mi(s))

def mf(f, s):
    return map(f, s)

def lmf(f, s):
    return list(mf(f, s))

def js(lst):
    return " ".join(str(d) for d in lst)

def jsns(lst):
    return "".join(str(d) for d in lst)

def line():
    return sys.stdin.readline().strip()

def linesp():
    return line().split()

def iline():
    return int(line())

def mat(n):
    matr = []
    for _ in range(n):
        matr.append(linesp())
    return matr

def matns(n):
    mat = []
    for _ in range(n):
        mat.append([c for c in line()])
    return mat

def mati(n):
    mat = []
    for _ in range(n):
       mat.append(lmi(line())) 
    return mat

def pmat(mat):
    for row in mat:
        print(js(row))

def main():
    line()
    n=[c for c in line()]
    new=[]
    # Assume we can get a number d1d2d3
    # such that it can't be divided anymore.
    # Assume that we can still get a larger number.
    for d in n:
        if d=='0' or d=='1':
            continue
        elif d=='4':
            new.append(3)
            new.append(2)
            new.append(2)
        elif d=='6':
            new.append(5)
            new.append(3)
        elif d=='8':
            new.append(7)
            new.append(2)
            new.append(2)
            new.append(2)
        elif d=='9':
            new.append(7)
            new.append(3)
            new.append(3)
            new.append(2)
        else:
            new.append(int(d))
    new.sort(reverse=True)
    print(jsns(new))
main()
