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 A {
	static InputStreamReader isr = new InputStreamReader(System.in);
	static BufferedReader br = new BufferedReader(isr);

	static int[] readIntArray() throws IOException {
		String[] v = br.readLine().split(" ");
		int[] ans = new int[v.length];

		for (int i = 0; i < ans.length; i++) {
			ans[i] = Integer.valueOf(v[i]);
		}

		return ans;
	}

	public static void main(String[] args) throws IOException {
		int n = readIntArray()[0];
		
		char[] num = br.readLine().toCharArray();
		int[] cifers = new int[10];
		for (int i = 0; i < num.length; i++) {
			int c = num[i] - '0';
			if(c > 1) {
				for (int j = c; j >= 2; j--) {
					int cc = j;
					while(cc > 1) {
						for (int k = 2; k <= 9; k++) {
							if (cc % k == 0) {
								cc /= k;
								cifers[k]++;
								break;
							}
						}
					}
				}				
			}			
		}
		
		StringBuilder ans = new StringBuilder();
		
		for (int i = 9; i >= 2; i--) {
			int t = cifers[i];
			for (int j = 0; j < t; j++) {
				
				dec(cifers, i);
				
				ans.append(i);	
			}			
		}
		
		System.out.println(ans.toString());
	}
	
	static void dec(int[] c, int n) {		
		
		for (int j = n; j >= 2; j--) {
			int cc = j;
			while(cc > 1) {
				for (int k = 2; k <= 9; k++) {
					if (cc % k == 0) {
						cc /= k;
						c[k]--;
					}
				}
			}
		}
	}
	
}
