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.*;
import java.util.*;

public class a{
	public static void main(String args[]){
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		String s = in.next();
		int a[] = new int[s.length()];
		for (int i = 0; i < n; i++)
			a[i] = (int)s.charAt(i) - (int)'0';
		ArrayList<Integer> ans = new ArrayList<Integer>();
		for (int i = 0; i < n; i++)
			switch(s.charAt(i)){
			case '0': case '1':
				break;
			case '2': case '3': case '5' : case '7':
				ans.add((int)s.charAt(i) - (int)'0');
				break;
			case '4':
				ans.add(3);
				ans.add(2);
				ans.add(2);
				break;
			case '6':
				ans.add(5);
				ans.add(3);
				break;
			case '8':
				ans.add(7);
				ans.add(2);
				ans.add(2);
				ans.add(2);
				break;
			case '9':
				ans.add(7);
				ans.add(3);
				ans.add(3);
				ans.add(2);
				break;
			}
		Collections.sort(ans, new Comparator<Integer>(){
			@Override
			public int compare(Integer a, Integer b){
				return b - a;
			}
		});
		for (int i = 0; i < ans.size(); i++)
			System.out.print(ans.get(i));
		System.out.println();
	}
}

