#include <bits/stdc++.h>
using namespace std;
int main() {
  string a;
  int n, i;
  cin >> n >> a;
  vector<int> v;
  for (i = 0; i < n; i++) {
    switch (a[i]) {
      case '0':
        break;
      case '1':
        break;
      case '4':
        v.push_back(3);
        v.push_back(2);
        v.push_back(2);
        break;
      case '6':
        v.push_back(5);
        v.push_back(3);
        break;
      case '8':
        v.push_back(7);
        v.push_back(2);
        v.push_back(2);
        v.push_back(2);
        break;
      case '9':
        v.push_back(7);
        v.push_back(3);
        v.push_back(3);
        v.push_back(2);
        break;
      default:
        v.push_back(a[i] - '0');
    }
  }
  sort(v.rbegin(), v.rend());
  for (i = 0; i < v.size(); i++) cout << v[i];
  return 0;
}

### Prompt

Create a solution in Cpp for 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 a;
  int n, i;
  cin >> n >> a;
  vector<int> v;
  for (i = 0; i < n; i++) {
    switch (a[i]) {
      case '0':
        break;
      case '1':
        break;
      case '4':
        v.push_back(3);
        v.push_back(2);
        v.push_back(2);
        break;
      case '6':
        v.push_back(5);
        v.push_back(3);
        break;
      case '8':
        v.push_back(7);
        v.push_back(2);
        v.push_back(2);
        v.push_back(2);
        break;
      case '9':
        v.push_back(7);
        v.push_back(3);
        v.push_back(3);
        v.push_back(2);
        break;
      default:
        v.push_back(a[i] - '0');
    }
  }
  sort(v.rbegin(), v.rend());
  for (i = 0; i < v.size(); i++) cout << v[i];
  return 0;
}

```