#include <bits/stdc++.h>
using namespace std;
multiset<int, greater<int> > sets;
multiset<int, greater<int> >::iterator p;
int main() {
  int n;
  char ch;
  while (scanf("%d\n", &n) != EOF) {
    sets.erase(sets.begin(), sets.end());
    while (n--) {
      ch = getchar();
      switch (ch) {
        case '0':
          break;
        case '1':
          break;
        case '4':
          sets.insert(3);
          sets.insert(2);
          sets.insert(2);
          break;
        case '6':
          sets.insert(3);
          sets.insert(5);
          break;
        case '8':
          sets.insert(7);
          sets.insert(2);
          sets.insert(2);
          sets.insert(2);
          break;
        case '9':
          sets.insert(7);
          sets.insert(3);
          sets.insert(3);
          sets.insert(2);
          break;
        default:
          sets.insert(ch - '0');
          break;
      }
    }
    ch = getchar();
    for (p = sets.begin(); p != sets.end(); ++p) printf("%d", *p);
    printf("\n");
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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;
multiset<int, greater<int> > sets;
multiset<int, greater<int> >::iterator p;
int main() {
  int n;
  char ch;
  while (scanf("%d\n", &n) != EOF) {
    sets.erase(sets.begin(), sets.end());
    while (n--) {
      ch = getchar();
      switch (ch) {
        case '0':
          break;
        case '1':
          break;
        case '4':
          sets.insert(3);
          sets.insert(2);
          sets.insert(2);
          break;
        case '6':
          sets.insert(3);
          sets.insert(5);
          break;
        case '8':
          sets.insert(7);
          sets.insert(2);
          sets.insert(2);
          sets.insert(2);
          break;
        case '9':
          sets.insert(7);
          sets.insert(3);
          sets.insert(3);
          sets.insert(2);
          break;
        default:
          sets.insert(ch - '0');
          break;
      }
    }
    ch = getchar();
    for (p = sets.begin(); p != sets.end(); ++p) printf("%d", *p);
    printf("\n");
  }
  return 0;
}

```