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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class ProblemA {
    BufferedReader rd;

    private ProblemA() throws IOException {
        rd = new BufferedReader(new InputStreamReader(System.in));
        compute();
    }

    private void compute() throws IOException {
        rd.readLine();
        char[] s = rd.readLine().toCharArray();
        int[] cnt = new int[10];
        for(char c: s) {
            switch(c) {
                case '2':
                    cnt[2]++;
                    break;
                case '3':
                    cnt[3]++;
                    break;
                case '4':
                    cnt[2] += 2;
                    cnt[3]++;
                    break;
                case '5':
                    cnt[5]++;
                    break;
                case '6':
                    cnt[3]++;
                    cnt[5]++;
                    break;
                case '7':
                    cnt[7]++;
                    break;
                case '8':
                    cnt[7]++;
                    cnt[2] += 3;
                    break;
                case '9':
                    cnt[2]++;
                    cnt[3] += 2;
                    cnt[7]++;
                    break;
            }
        }
        StringBuilder buf = new StringBuilder();
        for(int i=cnt.length-1;i>=0;i--) {
            for(int j=0;j<cnt[i];j++) {
                buf.append(i);
            }
        }
        out(buf);
    }

    private static void out(Object x) {
        System.out.println(x);
    }

    public static void main(String[] args) throws IOException {
        new ProblemA();
    }
}
