#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  vector<vector<int>> divisors{{},  {},     {2}, {3},          {2, 2, 3},
                               {5}, {5, 3}, {7}, {7, 2, 2, 2}, {7, 3, 3, 2}};
  int n;
  cin >> n;
  string s;
  cin >> s;
  vector<int> d;
  for (auto c : s) {
    auto digit = c - '0';
    if (digit >= 2) {
      for (auto j : divisors[digit]) {
        d.push_back(j);
      }
    }
  }
  sort(d.rbegin(), d.rend());
  for (auto i : d) {
    cout << i;
  }
}

### Prompt

Develop a solution in cpp 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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  vector<vector<int>> divisors{{},  {},     {2}, {3},          {2, 2, 3},
                               {5}, {5, 3}, {7}, {7, 2, 2, 2}, {7, 3, 3, 2}};
  int n;
  cin >> n;
  string s;
  cin >> s;
  vector<int> d;
  for (auto c : s) {
    auto digit = c - '0';
    if (digit >= 2) {
      for (auto j : divisors[digit]) {
        d.push_back(j);
      }
    }
  }
  sort(d.rbegin(), d.rend());
  for (auto i : d) {
    cout << i;
  }
}

```