#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int a[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t, n, k = 0;
  string s, r;
  cin >> t;
  cin >> s;
  for (int i = 0; i < s.length(); i++) {
    k = s[i] - '0';
    if (k % 2 == 0 && k > 2) {
      if (k == 6)
        r = r + '5' + '3';
      else if (k == 4)
        r = r + '3' + '2' + '2';
      else if (k == 8)
        r = r + '7' + '2' + '2' + '2';
    } else if (k == 9)
      r = r + '7' + '3' + '3' + '2';
    else if (k > 1)
      r += s[i];
  }
  sort(r.rbegin(), r.rend());
  cout << r << endl;
  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;
const int N = 2e5 + 10;
int a[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t, n, k = 0;
  string s, r;
  cin >> t;
  cin >> s;
  for (int i = 0; i < s.length(); i++) {
    k = s[i] - '0';
    if (k % 2 == 0 && k > 2) {
      if (k == 6)
        r = r + '5' + '3';
      else if (k == 4)
        r = r + '3' + '2' + '2';
      else if (k == 8)
        r = r + '7' + '2' + '2' + '2';
    } else if (k == 9)
      r = r + '7' + '3' + '3' + '2';
    else if (k > 1)
      r += s[i];
  }
  sort(r.rbegin(), r.rend());
  cout << r << endl;
  return 0;
}

```