#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  map<int, int> mm;
  for (int i = 0; i < s.length(); i++) {
    int xx = s[i] - '0';
    if (xx > 1) {
      if (xx == 4) {
        mm[2] += 2;
        mm[3]++;
      } else if (xx == 8) {
        mm[2] += 3;
        mm[7]++;
      } else if (xx == 6) {
        mm[3]++;
        mm[5]++;
      } else if (xx == 9) {
        mm[3] += 2;
        mm[2]++;
        mm[7]++;
      } else {
        mm[xx]++;
      }
    }
  }
  map<int, int>::reverse_iterator it;
  for (it = mm.rbegin(); it != mm.rend(); it++) {
    int e = it->second;
    while (e--) {
      cout << it->first;
    }
  }
  return 0;
}

### Prompt

Please formulate 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() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  map<int, int> mm;
  for (int i = 0; i < s.length(); i++) {
    int xx = s[i] - '0';
    if (xx > 1) {
      if (xx == 4) {
        mm[2] += 2;
        mm[3]++;
      } else if (xx == 8) {
        mm[2] += 3;
        mm[7]++;
      } else if (xx == 6) {
        mm[3]++;
        mm[5]++;
      } else if (xx == 9) {
        mm[3] += 2;
        mm[2]++;
        mm[7]++;
      } else {
        mm[xx]++;
      }
    }
  }
  map<int, int>::reverse_iterator it;
  for (it = mm.rbegin(); it != mm.rend(); it++) {
    int e = it->second;
    while (e--) {
      cout << it->first;
    }
  }
  return 0;
}

```