#include <bits/stdc++.h>
using namespace std;
map<char, string> mymap;
void setmap() {
  mymap['0'] = "";
  mymap['1'] = "";
  mymap['2'] = "2";
  mymap['3'] = "3";
  mymap['4'] = "322";
  mymap['5'] = "5";
  mymap['6'] = "53";
  mymap['7'] = "7";
  mymap['8'] = "7222";
  mymap['9'] = "7332";
}
int main() {
  int n;
  string s;
  cin >> n;
  cin >> s;
  setmap();
  string s1 = mymap[s[0]];
  for (int i = 1; i < s.length(); i++) {
    s1 += mymap[s[i]];
  }
  sort(s1.rbegin(), s1.rend());
  cout << s1 << endl;
  return 0;
}

### Prompt

Generate 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<char, string> mymap;
void setmap() {
  mymap['0'] = "";
  mymap['1'] = "";
  mymap['2'] = "2";
  mymap['3'] = "3";
  mymap['4'] = "322";
  mymap['5'] = "5";
  mymap['6'] = "53";
  mymap['7'] = "7";
  mymap['8'] = "7222";
  mymap['9'] = "7332";
}
int main() {
  int n;
  string s;
  cin >> n;
  cin >> s;
  setmap();
  string s1 = mymap[s[0]];
  for (int i = 1; i < s.length(); i++) {
    s1 += mymap[s[i]];
  }
  sort(s1.rbegin(), s1.rend());
  cout << s1 << endl;
  return 0;
}

```