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.
inp = '1234'
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]:
        output.append(2)
        if local//2 in [2,3]:
          output.append(local//2)
        local = local//2
        print(j, output)
    elif j % 3 == 0:
      local = j
      while local not in [0, 1, 2, 3]:
        output.append(3)
        if local//3 in [2,3]:
          output.append(local//3)
        local = local//3
        print(j, output)
    j -= 1
print(sorted(output, reverse=True))
    


