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 java.util.Arrays;
import java.util.HashMap;
import java.util.Scanner;

/**
 * Created by baktyzhan on 10.03.2015.
 */
public class FactorialP {
    public static void main(String[] args) {
        HashMap<Integer, String> map = new HashMap<Integer, String>() {
            {
                put(0, "");
                put(1, "");
                put(2, "2");
                put(3, "3");
                put(4, "223");
                put(5, "5");
                put(6, "35");
                put(7, "7");
                put(8, "7222");
                put(9, "7332");
            }
        };

        Scanner in = new Scanner(System.in);

        int n = Integer.parseInt(in.nextLine());
        String l = in.nextLine();

        String r = "";
        for (int i = 0; i < l.length(); i++) {
            r += map.get(Integer.parseInt(l.charAt(i) + ""));
        }
        char[] chars = r.toCharArray();
        Arrays.sort(chars);
        StringBuilder reverse = new StringBuilder(new String(chars)).reverse();
        System.out.println(reverse.toString());

    }


}
