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>
//Code by Sounak, IIEST
import java.io.*;
import java.math.*;
import java.util.*;
import java.util.Arrays;
public class Test1{
    public static void main(String args[])throws IOException{
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        char ch[]=sc.next().toCharArray();
        int a[]=new int[225];
        int i,j,a2=0,a3=0,a5=0,a7=0;
        for(i=0;i<n;i++)
        {
            switch (ch[i])
            {
                case '1':
                    break;
                case '0':
                    break;
                case '2':
                    a2++;
                    break;
                case '3':
                    a3++;
                    break;
                case '4':
                    a2+=2;
                    a3++;
                    break;
                case '5':
                    a5++;
                    break;
                case '6':
                    a3+=1;
                    a5+=1;
                    break;
                case '7':
                    a7++;
                    break;
                case '8':
                    a2+=3;
                    a7++;
                    break;
                case '9':
                    a2+=1;
                    a3+=2;
                    a7++;
                    break;
                
            }
        }
        for(i=0;i<a7;i++)
        System.out.print(7);
        for(i=0;i<a5;i++)
        System.out.print(5);
        for(i=0;i<a3;i++)
        System.out.print(3);
        for(i=0;i<a2;i++)
        System.out.print(2);
        System.out.println();
    }
}