#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n;
  cin >> n;
  vector<string> v(10);
  v[2] = "2";
  v[3] = "3";
  v[4] = "322";
  v[5] = "5";
  v[6] = "35";
  v[7] = "7";
  v[8] = "7222";
  v[9] = "7332";
  string s;
  cin >> s;
  string ans;
  for (auto x : s) {
    if (x != '0' && x != '1') {
      long long t = x - '0';
      ans = ans + v[t];
    }
  }
  sort(ans.rbegin(), ans.rend());
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
void solve() {
  long long n;
  cin >> n;
  vector<string> v(10);
  v[2] = "2";
  v[3] = "3";
  v[4] = "322";
  v[5] = "5";
  v[6] = "35";
  v[7] = "7";
  v[8] = "7222";
  v[9] = "7332";
  string s;
  cin >> s;
  string ans;
  for (auto x : s) {
    if (x != '0' && x != '1') {
      long long t = x - '0';
      ans = ans + v[t];
    }
  }
  sort(ans.rbegin(), ans.rend());
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) solve();
  return 0;
}

```