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;
int main() {
  int n, cnt = 0;
  scanf("%d", &n);
  char a[n + 1], res[61];
  scanf("%s", a);
  for (int i = 0; i < n; i++) {
    if (a[i] == '4' || a[i] == '8') {
      res[cnt++] = a[i] - 1;
      int temp = a[i] - '0';
      while (temp > 1) {
        res[cnt++] = '2';
        temp >>= 1;
      }
    } else if (a[i] == '6') {
      res[cnt++] = '3';
      res[cnt++] = '5';
    } else if (a[i] == '9') {
      res[cnt++] = '3';
      res[cnt++] = '3';
      res[cnt++] = '2';
      res[cnt++] = '7';
    } else if (a[i] != '0' && a[i] != '1') {
      res[cnt++] = a[i];
    }
  }
  sort(res, res + cnt);
  for (int i = cnt - 1; i >= 0; i--) {
    printf("%c", res[i]);
  }
  return 0;
}
