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 a1 {
	public static void main(String[] args) throws FileNotFoundException {
		Scanner in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		int n = in.nextInt();
		String s = in.next();
		int count = 0;
		ArrayList<Integer>[] a = new ArrayList[9];
		for (int i = 0; i < 9; i++) {
			a[i] = new ArrayList<Integer>();
		}
		for (int i = 0; i < n; i++) {
			int x = (int)s.charAt(i) - (int) '0';
			if (x >= 9){
				a[3].add(3);
				a[3].add(3);
			}
			if (x >= 8){
				a[2].add(2);
				a[2].add(2);
				a[2].add(2);
			}
			if (x >= 7){
				a[7].add(7);
			}
			if (x >= 6){
				a[2].add(2);
				a[3].add(3);
			}
			if (x >= 5){
				a[5].add(5);
			}
			if (x >= 4){
				a[2].add(2);
				a[2].add(2);
			}
			if (x >= 3){
				a[3].add(3);
			}
			if (x >= 2){
				a[2].add(2);
			}
		}
		ArrayList<Integer> ans = new ArrayList<Integer>();
		for (int i = 7; i > 1; i--) {
			while (a[i].size() > 0){
				for (int j = 2; j <= i; j++) {
					if (j == 2 || j == 3 || j == 5 || j == 7){
						int cnt = i / j + i / (j * j) + i / (j * j * j);
						for (int k = 0; k < cnt; k++) {
							a[j].remove(0);	
						}
					}
				}
				ans.add(i);
			}
		}
		for (int i = 0; i < ans.size(); i++) {
			out.print(ans.get(i));
		}
		out.close();
	}
}
