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>
#include <bits/stdc++.h>
using namespace std;
char a[105], c[200000];
int main() {
  int m;
  scanf("%d", &m);
  getchar();
  gets(a);
  char ch;
  int flag = 0;
  int i, j = 0;
  int l = strlen(a);
  for (i = 0; i < l; i++) {
    if (a[i] == '0' || a[i] == '1') continue;
    if (a[i] == '2' || a[i] == '3' || a[i] == '5' || a[i] == '7')
      c[flag++] = a[i];
    else if (a[i] == '4') {
      c[flag++] = '2';
      c[flag++] = '2';
      c[flag++] = '3';
    } else if (a[i] == '8') {
      c[flag++] = '2';
      c[flag++] = '2';
      c[flag++] = '2';
      c[flag++] = '7';
    } else if (a[i] == '6') {
      c[flag++] = '3';
      c[flag++] = '5';
    } else if (a[i] == '9') {
      c[flag++] = '3';
      c[flag++] = '3';
      c[flag++] = '2';
      c[flag++] = '7';
    }
  }
  c[flag] = '\0';
  sort(c, c + flag);
  l = strlen(c);
  for (i = l - 1; i >= 0; i--) cout << c[i];
  cout << endl;
  return 0;
}
