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 input[111];
int N;
int main(int argc, char *argv[]) {
  scanf("%d", &N);
  scanf("%s", input);
  vector<int> ans;
  for (int ii = 0; ii < (N); ii++) {
    int i = input[ii] - '0';
    if (i < 2) continue;
    if (i == 4) {
      ans.push_back(3);
      ans.push_back(2);
      ans.push_back(2);
    } else if (i == 6) {
      ans.push_back(5);
      ans.push_back(3);
    } else if (i == 8) {
      ans.push_back(7);
      ans.push_back(2);
      ans.push_back(2);
      ans.push_back(2);
    } else if (i == 9) {
      ans.push_back(7);
      ans.push_back(3);
      ans.push_back(3);
      ans.push_back(2);
    } else
      ans.push_back(i);
  }
  sort(ans.begin(), ans.end());
  reverse(ans.begin(), ans.end());
  for (int i = 0; i < (ans.size()); i++) printf("%d", ans[i]);
  printf("\n");
  return 0;
}
