#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int n, m, ar[60], br[60], i = 0, j, k = 0;
  cin >> m;
  cin >> n;
  while (n) {
    if (n % 10 != 0 && n % 10 != 1) {
      ar[i] = n % 10;
      i++;
    }
    n = n / 10;
  }
  for (j = 0; j < i; j++) {
    if (ar[j] == 9) {
      br[k++] = 7;
      br[k++] = 3;
      br[k++] = 3;
      br[k++] = 2;
    } else if (ar[j] == 2 || (ar[j] % 2 == 1)) {
      br[k++] = ar[j];
    } else {
      br[k++] = ar[j] - 1;
      if (ar[j] == 6) {
        br[k++] = 3;
      } else {
        br[k++] = 2;
        br[k++] = 2;
        if (ar[j] == 8) br[k++] = 2;
      }
    }
  }
  sort(br, br + k);
  for (i = k - 1; i >= 0; i--) cout << br[i];
  cout << endl;
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution 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;
int main() {
  long long int n, m, ar[60], br[60], i = 0, j, k = 0;
  cin >> m;
  cin >> n;
  while (n) {
    if (n % 10 != 0 && n % 10 != 1) {
      ar[i] = n % 10;
      i++;
    }
    n = n / 10;
  }
  for (j = 0; j < i; j++) {
    if (ar[j] == 9) {
      br[k++] = 7;
      br[k++] = 3;
      br[k++] = 3;
      br[k++] = 2;
    } else if (ar[j] == 2 || (ar[j] % 2 == 1)) {
      br[k++] = ar[j];
    } else {
      br[k++] = ar[j] - 1;
      if (ar[j] == 6) {
        br[k++] = 3;
      } else {
        br[k++] = 2;
        br[k++] = 2;
        if (ar[j] == 8) br[k++] = 2;
      }
    }
  }
  sort(br, br + k);
  for (i = k - 1; i >= 0; i--) cout << br[i];
  cout << endl;
  return 0;
}

```