#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, flag = 0;
  cin >> n;
  char num[n];
  string ne = "";
  vector<int> vec;
  cin >> ne;
  for (int i = 0; i < n; i++) {
    num[i] = ne[i] - '0';
    if (num[i] == 3 || num[i] == 5 || num[i] == 7 || num[i] == 2)
      vec.push_back(num[i]);
    else if (num[i] == 4) {
      vec.push_back(3);
      vec.push_back(2);
      vec.push_back(2);
    } else if (num[i] == 6) {
      vec.push_back(5);
      vec.push_back(3);
    } else if (num[i] == 8) {
      vec.push_back(7);
      vec.push_back(2);
      vec.push_back(2);
      vec.push_back(2);
    } else if (num[i] == 9) {
      vec.push_back(7);
      vec.push_back(3);
      vec.push_back(3);
      vec.push_back(2);
    }
  }
  sort(vec.begin(), vec.end());
  for (int i = 0; i < vec.size(); i++) cout << vec[vec.size() - i - 1];
  return 0;
}

### 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() {
  int n, flag = 0;
  cin >> n;
  char num[n];
  string ne = "";
  vector<int> vec;
  cin >> ne;
  for (int i = 0; i < n; i++) {
    num[i] = ne[i] - '0';
    if (num[i] == 3 || num[i] == 5 || num[i] == 7 || num[i] == 2)
      vec.push_back(num[i]);
    else if (num[i] == 4) {
      vec.push_back(3);
      vec.push_back(2);
      vec.push_back(2);
    } else if (num[i] == 6) {
      vec.push_back(5);
      vec.push_back(3);
    } else if (num[i] == 8) {
      vec.push_back(7);
      vec.push_back(2);
      vec.push_back(2);
      vec.push_back(2);
    } else if (num[i] == 9) {
      vec.push_back(7);
      vec.push_back(3);
      vec.push_back(3);
      vec.push_back(2);
    }
  }
  sort(vec.begin(), vec.end());
  for (int i = 0; i < vec.size(); i++) cout << vec[vec.size() - i - 1];
  return 0;
}

```