#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int n, i;
  cin >> n;
  string s;
  cin >> s;
  map<char, long long int> m;
  vector<long long int> v;
  m['2'] = 2;
  m['3'] = 3;
  m['4'] = 322;
  m['5'] = 5;
  m['6'] = 53;
  m['7'] = 7;
  m['8'] = 7222;
  m['9'] = 7332;
  long long int x, r;
  for (i = 0; i < n; i++) {
    if (s[i] != '0' && s[i] != '1') {
      x = m[s[i]];
      while (x > 0) {
        r = x % 10;
        v.push_back(r);
        x /= 10;
      }
    }
  }
  sort(v.begin(), v.end(), greater<long long int>());
  for (i = 0; i < v.size(); i++) {
    cout << v[i];
  }
  cout << endl;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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() {
  long long int n, i;
  cin >> n;
  string s;
  cin >> s;
  map<char, long long int> m;
  vector<long long int> v;
  m['2'] = 2;
  m['3'] = 3;
  m['4'] = 322;
  m['5'] = 5;
  m['6'] = 53;
  m['7'] = 7;
  m['8'] = 7222;
  m['9'] = 7332;
  long long int x, r;
  for (i = 0; i < n; i++) {
    if (s[i] != '0' && s[i] != '1') {
      x = m[s[i]];
      while (x > 0) {
        r = x % 10;
        v.push_back(r);
        x /= 10;
      }
    }
  }
  sort(v.begin(), v.end(), greater<long long int>());
  for (i = 0; i < v.size(); i++) {
    cout << v[i];
  }
  cout << endl;
  return 0;
}

```