#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 10;
int main() {
  ios_base::sync_with_stdio(false);
  map<char, string> mp = {
      {'0', "0"}, {'1', "0"},  {'2', "2"}, {'3', "3"},    {'4', "223"},
      {'5', "5"}, {'6', "35"}, {'7', "7"}, {'8', "2227"}, {'9', "2337"},
  };
  int n;
  string str, ans = "";
  cin >> n >> str;
  for (int i = 0; i < n; i++) ans += mp[str[i]];
  vector<int> vec;
  for (int i = 0; i < ans.length(); i++) {
    int curr = ans[i] - '0';
    if (curr != 0) vec.push_back(curr);
  }
  sort(vec.begin(), vec.end(), greater<int>());
  for (auto l : vec) cout << l;
}

### 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;
const int inf = 1e9 + 10;
int main() {
  ios_base::sync_with_stdio(false);
  map<char, string> mp = {
      {'0', "0"}, {'1', "0"},  {'2', "2"}, {'3', "3"},    {'4', "223"},
      {'5', "5"}, {'6', "35"}, {'7', "7"}, {'8', "2227"}, {'9', "2337"},
  };
  int n;
  string str, ans = "";
  cin >> n >> str;
  for (int i = 0; i < n; i++) ans += mp[str[i]];
  vector<int> vec;
  for (int i = 0; i < ans.length(); i++) {
    int curr = ans[i] - '0';
    if (curr != 0) vec.push_back(curr);
  }
  sort(vec.begin(), vec.end(), greater<int>());
  for (auto l : vec) cout << l;
}

```