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

### Prompt

Generate 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;
vector<int> numbers;
vector<int> ans;
int main(int argc, char const *argv[]) {
  int n;
  cin >> n;
  string s;
  cin >> s;
  for (int i = 0; i < n; i++) {
    if ((s[i] - '0') > 1) numbers.push_back(s[i] - '0');
  }
  for (int i = 0; i < numbers.size(); i++) {
    switch (numbers[i]) {
      case 2:
        ans.push_back(2);
        break;
      case 3:
        ans.push_back(3);
        break;
      case 4:
        ans.push_back(2);
        ans.push_back(2);
        ans.push_back(3);
        break;
      case 5:
        ans.push_back(5);
        break;
      case 6:
        ans.push_back(5);
        ans.push_back(3);
        break;
      case 7:
        ans.push_back(7);
        break;
      case 8:
        ans.push_back(7);
        ans.push_back(2);
        ans.push_back(2);
        ans.push_back(2);
        break;
      case 9:
        ans.push_back(7);
        ans.push_back(2);
        ans.push_back(3);
        ans.push_back(3);
        break;
    }
  }
  sort(ans.rbegin(), ans.rend());
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i];
  }
  cout << endl;
  return 0;
}

```