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 math
n = input()
inp = input()
i = 0
while i < len(inp):
  if inp[i] != '0':
    break
  i += 1
num = inp[i:]
output = []
primes = [2,3,5,7]
output = []
for i in range(len(num)):
  c = int(num[i])
  j = c
  while j >= 0:
    if j in primes:
      # print('char', c)
      # print('prime', j)
      output.append(j)
      break
    if j % 2 == 0:
      local = j
      while local not in [0, 1, 2, 3]:
        if local//2 == 2:
          output.append(2)
          output.append(local//2)
        elif local//2 == 3:
          output.append(3)
        else:
          output.append(2)
        local = local//2
        # print(j, output)
    elif j % 3 == 0:
      local = j
      if j == 9:
        output.append(3)
        output.append(3)
        output.append(2)
        j -=1 
      else:
        while local not in [0, 1, 2, 3]:
          if local//3 == 2:
            output.append(3)
            output.append(local//3)
          elif local//3 == 3:
            output.append(3)
          else:
            output.append(3)
          local = local//3
        # print(j, output)
    j -= 1
ans = ''.join(str(x) for x in sorted(output, reverse=True))
print(ans)
# factorial = 1
# for c in num:
#   factorial *= math.factorial(int(c))
# print(factorial)
# factorial = 1
# for c in ans:
#   factorial *= math.factorial(int(c))
# print(factorial)



    


