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;

/*
4
1234

3
555

1
9

 */

public class c {

	public static void main(String[] args)
	{
		Scanner in = new Scanner(System.in);
		
		in.nextInt();
		String number = in.next();
		
		in.close();
		
		int[] counts = new int[10];
		for(char c: number.toCharArray())
		{
			int num = c - '0';
			
			for(int curr = 2; curr <= num; ++curr)
			{
				++counts[curr];
			}
		}
		boolean[] primes = new boolean[10];
		primes[2] = primes[3] = primes[5] = primes[7] = true;
		
		ArrayList<Integer> digits = new ArrayList<Integer>();

		//System.out.println(Arrays.toString(counts));
		
		for(int curr = 9; curr >= 2; --curr)
		{
			while(primes[curr] && counts[curr] > 0)
			{
				digits.add(curr);
				for(int digit = curr; digit >= 2; -- digit)
				{
					--counts[digit];
				}
			}
		}
		
		//System.out.println(Arrays.toString(counts));
		
		counts[3] += 2*counts[9] + counts[6];
		counts[2] += 3*counts[8] + counts[6] + 2*counts[4];
		counts[9] = 0;
		counts[8] = 0;
		counts[6] = 0;
		counts[4] = 0;
		
		
		for(int curr = 9; curr >= 2; --curr)
		{
			while(counts[curr] > 0)
			{
				digits.add(curr);
				for(int digit = curr; digit >= 2; -- digit)
				{
					--counts[digit];
				}
			}
		}
		
		Collections.sort(digits, Collections.reverseOrder());
		for(int digit: digits)
			System.out.print(digit);
		
		System.out.println();
		
	}
}
