#include <bits/stdc++.h>
using namespace std;
int main() {
  string s, ns;
  int n, i, num;
  while (cin >> n >> s) {
    for (i = 0; i < n; i++) {
      if (s[i] != '0' && s[i] != '1') ns += s[i];
    }
    n = ns.size();
    s.clear();
    for (i = 0; i < n; i++) {
      if (ns[i] == '4' || ns[i] == '8' || ns[i] == '6' || ns[i] == '9') {
        if (ns[i] == '4') {
          s += '3', s += '2', s += '2';
        } else if (ns[i] == '6') {
          s += '5', s += '3';
        } else if (ns[i] == '8') {
          s += '7', s += '2', s += '2', s += '2';
        } else {
          s += '7', s += '3', s += '3', s += '2';
        }
      } else
        s += ns[i];
    }
    n = s.size();
    sort(s.begin(), s.end());
    for (i = n - 1; i >= 0; i--) {
      cout << s[i];
    }
    cout << endl;
    s.clear(), ns.clear();
  }
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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() {
  string s, ns;
  int n, i, num;
  while (cin >> n >> s) {
    for (i = 0; i < n; i++) {
      if (s[i] != '0' && s[i] != '1') ns += s[i];
    }
    n = ns.size();
    s.clear();
    for (i = 0; i < n; i++) {
      if (ns[i] == '4' || ns[i] == '8' || ns[i] == '6' || ns[i] == '9') {
        if (ns[i] == '4') {
          s += '3', s += '2', s += '2';
        } else if (ns[i] == '6') {
          s += '5', s += '3';
        } else if (ns[i] == '8') {
          s += '7', s += '2', s += '2', s += '2';
        } else {
          s += '7', s += '3', s += '3', s += '2';
        }
      } else
        s += ns[i];
    }
    n = s.size();
    sort(s.begin(), s.end());
    for (i = n - 1; i >= 0; i--) {
      cout << s[i];
    }
    cout << endl;
    s.clear(), ns.clear();
  }
  return 0;
}

```