#include <bits/stdc++.h>
using namespace std;
map<int, string> store;
int main() {
  int n;
  cin >> n;
  string str;
  cin >> str;
  store[2] = "2";
  store[3] = "3";
  store[4] = "322";
  store[5] = "5";
  store[6] = "53";
  store[7] = "7";
  store[8] = "7222";
  store[9] = "7332";
  vector<char> ans;
  for (int i = 0; i < n; i++) {
    if (str[i] != '0' && str[i] != '1') {
      for (int j = 0; j < store[str[i] - '0'].size(); j++) {
        ans.push_back(store[str[i] - '0'][j]);
      }
    }
  }
  sort(ans.begin(), ans.end(), greater<char>());
  for (auto i : ans) cout << i;
  return 0;
}

### Prompt

Your task is to create 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;
map<int, string> store;
int main() {
  int n;
  cin >> n;
  string str;
  cin >> str;
  store[2] = "2";
  store[3] = "3";
  store[4] = "322";
  store[5] = "5";
  store[6] = "53";
  store[7] = "7";
  store[8] = "7222";
  store[9] = "7332";
  vector<char> ans;
  for (int i = 0; i < n; i++) {
    if (str[i] != '0' && str[i] != '1') {
      for (int j = 0; j < store[str[i] - '0'].size(); j++) {
        ans.push_back(store[str[i] - '0'][j]);
      }
    }
  }
  sort(ans.begin(), ans.end(), greater<char>());
  for (auto i : ans) cout << i;
  return 0;
}

```