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.
#def cast(cre, typ): return type(typ)(map(cre, typ));
def solution():
    n = int(input()); a = input();

    delidx0, delidx1 = 0, 0;
    while delidx0 != -1 or delidx1 != -1:
        delidx0 = a.find("0");
        delidx1 = a.find("1");
        if delidx0 != -1: a = a[:delidx0] + a[delidx0+1:];
        if delidx1 != -1: a = a[:delidx1] + a[delidx1+1:];

    for i in range(len(a)):
        if a[i] == "4":
            a = a[:i] + "322" + a[i+1];
        elif a[i] == "6":
            a = a[:i] + "53" + a[i+1];
        elif a[i] == "8":
            a = a[:i] + "7222" + a[i+1];
    
    a = "".join([int(e) for e in a].sort()[::-1]);

    print(a);
solution();
