#include <bits/stdc++.h>
using namespace std;
string s;
int k, i, x[1000], j, n;
int main() {
  cin >> n;
  cin >> s;
  k = 1;
  for (i = 0; i < n; i++) {
    int m = s[i] - '0';
    if (m == 6) {
      x[k] = 5;
      x[k + 1] = 3;
      k += 1;
    }
    if (m == 8) {
      x[k] = 7;
      x[k + 1] = 2;
      x[k + 2] = 2;
      x[k + 3] = 2;
      k += 3;
    }
    if (m == 9) {
      x[k] = 7;
      x[k + 1] = 2;
      x[k + 2] = 3;
      x[k + 3] = 3;
      k += 3;
    }
    if (m == 4) {
      x[k] = 2;
      x[k + 1] = 2;
      x[k + 2] = 3;
      k += 2;
    }
    if (m == 2 || m == 3 || m == 7 || m == 5) x[k] = m;
    if (m == 1 || m == 0) k--;
    k++;
  }
  k--;
  sort(x + 1, x + k + 1);
  for (i = k; i >= 1; i--) cout << x[i];
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
string s;
int k, i, x[1000], j, n;
int main() {
  cin >> n;
  cin >> s;
  k = 1;
  for (i = 0; i < n; i++) {
    int m = s[i] - '0';
    if (m == 6) {
      x[k] = 5;
      x[k + 1] = 3;
      k += 1;
    }
    if (m == 8) {
      x[k] = 7;
      x[k + 1] = 2;
      x[k + 2] = 2;
      x[k + 3] = 2;
      k += 3;
    }
    if (m == 9) {
      x[k] = 7;
      x[k + 1] = 2;
      x[k + 2] = 3;
      x[k + 3] = 3;
      k += 3;
    }
    if (m == 4) {
      x[k] = 2;
      x[k + 1] = 2;
      x[k + 2] = 3;
      k += 2;
    }
    if (m == 2 || m == 3 || m == 7 || m == 5) x[k] = m;
    if (m == 1 || m == 0) k--;
    k++;
  }
  k--;
  sort(x + 1, x + k + 1);
  for (i = k; i >= 1; i--) cout << x[i];
  return 0;
}

```