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.io.FileNotFoundException;
import java.util.Scanner;

public class A {
    public static void main(String zargs[]) throws FileNotFoundException {

        Scanner scanner = new Scanner(System.in);

        int n = scanner.nextInt();
        scanner.nextLine();
        String s = scanner.nextLine();

        int a[] = new int[10];
        for (int i = 0; i < s.length(); i++) {

            int x = s.charAt(i) - '0';

            for (int j = 2; j <= x; j++) {
                if (j == 4)
                    a[2] += 2;
                else if (j == 6) {
                    a[2]++;
                    a[3]++;
                } else if (j == 8)
                    a[2] += 3;
                else if (j == 9)
                    a[3] += 2;
                else
                    a[j]++;
            }
        }

        StringBuilder sb = new StringBuilder();

        sb.append(repeat("7", a[7]));
        a[2] -= 4 * a[7];
        a[3] -= 2 * a[7];
        a[5] -= a[7];
        a[7] = 0;


        sb.append(repeat("5", a[5]));
        a[3] -= a[5];
        a[2] -= 3 * a[5];
        a[5] = 0;

        sb.append(repeat("3", a[3]));
        a[2] -= a[3];
        a[3] = 0;

        sb.append(repeat("2", a[2]));

        System.out.println(sb.toString());
    }

    private static String repeat(String s, int x) {
        StringBuilder sb = new StringBuilder();

        while (x-- > 0)
            sb.append(s);
        return sb.toString();
    }
}
