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 n;
int x, ans[100];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%1d", &x);
    if (x == 2)
      ans[++ans[0]] = 2;
    else if (x == 3)
      ans[++ans[0]] = 3;
    else if (x == 4)
      ans[++ans[0]] = 2, ans[++ans[0]] = 2, ans[++ans[0]] = 3;
    else if (x == 5)
      ans[++ans[0]] = 5;
    else if (x == 6)
      ans[++ans[0]] = 3, ans[++ans[0]] = 5;
    else if (x == 7)
      ans[++ans[0]] = 7;
    else if (x == 8)
      ans[++ans[0]] = 7, ans[++ans[0]] = 2, ans[++ans[0]] = 2,
      ans[++ans[0]] = 2;
    else if (x == 9)
      ans[++ans[0]] = 7, ans[++ans[0]] = 2, ans[++ans[0]] = 3,
      ans[++ans[0]] = 3;
  }
  sort(ans + 1, ans + ans[0] + 1);
  for (int i = ans[0]; i >= 1; i--) printf("%d", ans[i]);
  puts("");
  return 0;
}
