#include <bits/stdc++.h>
using namespace std;
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(ios::badbit | ios::failbit);
  int n;
  string s;
  cin >> n >> s;
  string res;
  for (auto c : s) {
    if (c >= '2') {
      if (c == '4') {
        res += "223";
      } else if (c == '6') {
        res += "35";
      } else if (c == '8') {
        res += "7222";
      } else if (c == '9') {
        res += "7233";
      } else {
        res += c;
      }
    }
  }
  sort(res.rbegin(), res.rend());
  cout << res << "\n";
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(ios::badbit | ios::failbit);
  int n;
  string s;
  cin >> n >> s;
  string res;
  for (auto c : s) {
    if (c >= '2') {
      if (c == '4') {
        res += "223";
      } else if (c == '6') {
        res += "35";
      } else if (c == '8') {
        res += "7222";
      } else if (c == '9') {
        res += "7233";
      } else {
        res += c;
      }
    }
  }
  sort(res.rbegin(), res.rend());
  cout << res << "\n";
  return 0;
}

```