#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int n, a, b;
  cin >> n >> a;
  vector<long long int> ans;
  while (a > 0) {
    long long int x = a % 10;
    if (x == 2 || x == 3 || x == 5 || x == 7)
      ans.push_back(x);
    else if (x == 4)
      ans.push_back(3), ans.push_back(2), ans.push_back(2);
    else if (x == 6)
      ans.push_back(3), ans.push_back(5);
    else if (x == 8)
      ans.push_back(7), ans.push_back(2), ans.push_back(2), ans.push_back(2);
    else if (x == 9)
      ans.push_back(7), ans.push_back(3), ans.push_back(3), ans.push_back(2);
    a /= 10;
  }
  sort(ans.rbegin(), ans.rend());
  for (int i = 0; i < ans.size(); i++) cout << ans[i];
  cout << endl;
  return 0;
}

### Prompt

Please create a solution in cpp 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() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int n, a, b;
  cin >> n >> a;
  vector<long long int> ans;
  while (a > 0) {
    long long int x = a % 10;
    if (x == 2 || x == 3 || x == 5 || x == 7)
      ans.push_back(x);
    else if (x == 4)
      ans.push_back(3), ans.push_back(2), ans.push_back(2);
    else if (x == 6)
      ans.push_back(3), ans.push_back(5);
    else if (x == 8)
      ans.push_back(7), ans.push_back(2), ans.push_back(2), ans.push_back(2);
    else if (x == 9)
      ans.push_back(7), ans.push_back(3), ans.push_back(3), ans.push_back(2);
    a /= 10;
  }
  sort(ans.rbegin(), ans.rend());
  for (int i = 0; i < ans.size(); i++) cout << ans[i];
  cout << endl;
  return 0;
}

```