#include <bits/stdc++.h>
using namespace std;
char in[20];
int main() {
  int n;
  scanf("%d%s", &n, in);
  vector<int> ans;
  for (int i = 0; i < n; i++)
    if (in[i] == '2' || in[i] == '3' || in[i] == '5' || in[i] == '7')
      ans.push_back(in[i]);
    else if (in[i] == '4')
      ans.push_back('3'), ans.push_back('2'), ans.push_back('2');
    else if (in[i] == '6')
      ans.push_back('5'), ans.push_back('3');
    else if (in[i] == '8')
      ans.push_back('7'), ans.push_back('2'), ans.push_back('2'),
          ans.push_back('2');
    else if (in[i] == '9')
      ans.push_back('7'), ans.push_back('3'), ans.push_back('3'),
          ans.push_back('2');
  sort(ans.begin(), ans.end());
  for (int i = ((int)(ans).size()) - 1; i >= 0; i--) putchar(ans[i]);
  puts("");
  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;
char in[20];
int main() {
  int n;
  scanf("%d%s", &n, in);
  vector<int> ans;
  for (int i = 0; i < n; i++)
    if (in[i] == '2' || in[i] == '3' || in[i] == '5' || in[i] == '7')
      ans.push_back(in[i]);
    else if (in[i] == '4')
      ans.push_back('3'), ans.push_back('2'), ans.push_back('2');
    else if (in[i] == '6')
      ans.push_back('5'), ans.push_back('3');
    else if (in[i] == '8')
      ans.push_back('7'), ans.push_back('2'), ans.push_back('2'),
          ans.push_back('2');
    else if (in[i] == '9')
      ans.push_back('7'), ans.push_back('3'), ans.push_back('3'),
          ans.push_back('2');
  sort(ans.begin(), ans.end());
  for (int i = ((int)(ans).size()) - 1; i >= 0; i--) putchar(ans[i]);
  puts("");
  return 0;
}

```