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.lang.StringBuilder;
import java.math.BigInteger;

public class Solution {
	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int n = s.nextInt();
		String ss = s.next();
		int[] a = new int[n];
		for(int i = 0; i < n; i++) a[i] = ss.charAt(i) - '0';
		int[] nums = new int[10]; // 0 to 9
	    for(int i = 0; i < n; i++) {
	    	if(a[i] == 2) nums[2]++;
	    	else if(a[i] == 3) nums[3]++;
	    	else if(a[i] == 4) {
	    		nums[2] += 2;
	    		nums[3]++;
	    	}
	    	else if(a[i] == 5) nums[5]++;
	    	else if(a[i] == 6) {
	    		nums[3]++;
	    		nums[5]++;
	    	}
	    	else if(a[i] == 7) nums[7]++;
	    	else if(a[i] == 8) {
	    		nums[2] += 3;
	    		nums[7]++;
	    	}
	    	else if(a[i] == 9) {
	    		nums[2]++;
	    		nums[3] += 2;
	    		nums[7]++;
	    	}
	    }
	    for(int i = 9; i >= 2; i--) {
	    	for(int j = 0; j < nums[i]; j++) System.out.print(i);
	    }
	}
}
