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.*;
import java.io.*;

public class Main {
	public static void main(String [] args) throws IOException{
		Scanner in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		int n = in.nextInt();
		String st = in.next();
		int a[] = new int[10];
		for (int i = 0; i < n; i++) {
			int m = new Integer("" + st.charAt(i));
			int k = 2;
			int s = 1;
			for (int j=2;j<=m;j++)
				s *= j;
			while (s > 1) {
				if (s % k == 0) {
					s/= k;
					a[k]++;
				} else {
					k++;
				}
			}	 
		}
		for (int i=0;i<10;i++) {
			//out.println(i + " " + a[i]);
		}
		for (int i = 7; i > 1; i--) {
			//out.println(i + " > " + a[i]);
			for(int j = 0; j < a[i]; j++)
				out.print(i);
			for (int j = i-1;j>1;j--) { 
				if (i == 7) {
					if (j == 2) a[j] -= 4 * a[i]; else
					if (j == 3) a[j] -= 2 * a[i]; else
					a[j] -= a[i];
				} else
				if (i == 5) {
					if (j == 2) a[j] -= 3 * a[i]; else
					a[j] -= a[i];
					//out.println(i + " " + j + " " + a[j]);
				} else
				if (i == 3) {
					a[j] -= a[i];
				}
			}
		}
		out.close();
	}
}