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.Scanner;

public class Prob01 {
    
    static long[] digits = new long[10];

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        in.nextInt();
        String x = in.next();
        for (int i = 0; i < x.length(); i++) {
            int digit = x.charAt(i) - '0';
            if (digit <= 1) continue;
            add(digit);
        }
        StringBuilder sb = new StringBuilder();
        for (int i = 7; i >= 2; i--) {
            char c = (char) ('0' + i);
            while (digits[i] > 0) {
                sb.append(c);
                digits[i]--;
            }
        }
        System.out.println(sb.toString());
    }

    private static void add(int digit) {
        if (digit <= 1) return;
        
        if (digit == 9) {
            digits[2]++;
            digits[3]+=2;
            digits[7]++;
        } else if (digit == 8) {
            digits[2]+=3;
            digits[7]++;
        } else if (digit == 6) {
            digits[3]++;
            digits[5]++;
        } else if (digit == 4) {
          digits[2]+=2;
          digits[3]++;
        } else {
            digits[digit]++;
        }
    }

}
