#include <bits/stdc++.h>
using namespace std;
map<int, string> ma;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  string s;
  cin >> n >> s;
  ma[4] = "223";
  ma[6] = "35";
  ma[8] = "2227";
  ma[9] = "3327";
  string ans;
  for (int i = 0; i < n; i++) {
    int d = s[i] - '0';
    if (d <= 1) continue;
    if (ma.find(d) == ma.end())
      ans += char(d + '0');
    else
      ans += ma[d];
  }
  sort(ans.begin(), ans.end());
  reverse(ans.begin(), ans.end());
  cout << ans;
  return 0;
}

### 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;
map<int, string> ma;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  string s;
  cin >> n >> s;
  ma[4] = "223";
  ma[6] = "35";
  ma[8] = "2227";
  ma[9] = "3327";
  string ans;
  for (int i = 0; i < n; i++) {
    int d = s[i] - '0';
    if (d <= 1) continue;
    if (ma.find(d) == ma.end())
      ans += char(d + '0');
    else
      ans += ma[d];
  }
  sort(ans.begin(), ans.end());
  reverse(ans.begin(), ans.end());
  cout << ans;
  return 0;
}

```