#include <bits/stdc++.h>
using namespace std;
long long n, temp, no;
vector<long long> out;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  cout << fixed;
  cout << setprecision(0);
  cin >> no;
  for (long long i = 0; i < n; i++) {
    temp = no % 10;
    no /= 10;
    if (temp == 2 || temp == 3 || temp == 5 || temp == 7)
      out.push_back(temp);
    else if (temp == 4) {
      out.push_back(2);
      out.push_back(2);
      out.push_back(3);
    } else if (temp == 6) {
      out.push_back(3);
      out.push_back(5);
    } else if (temp == 8) {
      out.push_back(2);
      out.push_back(2);
      out.push_back(2);
      out.push_back(7);
    } else if (temp == 9) {
      out.push_back(2);
      out.push_back(3);
      out.push_back(3);
      out.push_back(7);
    }
  }
  sort(out.begin(), out.end(), greater<long long>());
  for (long long i = 0; i < out.size(); i++) cout << out[i];
}

### 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;
long long n, temp, no;
vector<long long> out;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  cout << fixed;
  cout << setprecision(0);
  cin >> no;
  for (long long i = 0; i < n; i++) {
    temp = no % 10;
    no /= 10;
    if (temp == 2 || temp == 3 || temp == 5 || temp == 7)
      out.push_back(temp);
    else if (temp == 4) {
      out.push_back(2);
      out.push_back(2);
      out.push_back(3);
    } else if (temp == 6) {
      out.push_back(3);
      out.push_back(5);
    } else if (temp == 8) {
      out.push_back(2);
      out.push_back(2);
      out.push_back(2);
      out.push_back(7);
    } else if (temp == 9) {
      out.push_back(2);
      out.push_back(3);
      out.push_back(3);
      out.push_back(7);
    }
  }
  sort(out.begin(), out.end(), greater<long long>());
  for (long long i = 0; i < out.size(); i++) cout << out[i];
}

```