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.Scanner;

public class A {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] a = new int[10];
        String s = in.next();
        for (int i = 0; i < n; i++) {
            int x = Integer.parseInt(s.substring(i, i + 1));
            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]++;
                }
            }
        }
        for (int i = 7; i >= 0; i--) {
            for (int j = 0; j < a[i]; j++) {
                System.out.print(i);
                for (int k = 2; k < i; k++) {
                    if (k == 4) {
                        a[2] -= 2;
                    } else if (k == 6) {
                        a[2]--;
                        a[3]--;
                    } else if (k == 8) {
                        a[2] -= 3;
                    } else if (k == 9){
                        a[3] -= 2;
                    } else {
                        a[k]--;
                    }
                }
            }
        }
        in.close();
    }

}
