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.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class Codeforces{
	
	static ArrayList<Integer> s;
	
	public static void main(String[] args){		
		Scanner scan = new Scanner(System.in);
		int n = Integer.parseInt(scan.nextLine());
		String input = scan.nextLine();
		s = new ArrayList<>();
		for(int i = 0; i < n; i++){
			int current = input.charAt(i) - 48;
			add(current);
		}	
		Collections.sort(s);
		for(int i = s.size() - 1; i >= 0; i--) System.out.print(s.get(i));
	}
	
	public static void add(int n){
		if(n == 9){
			s.add(2);
			s.add(3);
			s.add(3);
			s.add(7);
		}
		else if(n == 8){
			s.add(2);
			s.add(2);
			s.add(2);
			s.add(7);
		}
		else if(n == 7){
			s.add(7);
		}
		else if(n == 6){
			s.add(3);
			s.add(5);
		}
		else if(n == 5){
			s.add(5);
		}
		else if(n == 4){
			s.add(2);
			s.add(2);
			s.add(3);
		}
		else if(n == 3){
			s.add(3);
		}
		else if(n == 2){
			s.add(2);
		}
	}
}








