#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  char vet[20];
  vector<char> res;
  scanf("%d", &n);
  scanf("%s", vet);
  int i;
  for (i = 0; i < n; i++) {
    if (vet[i] > '1') {
      if (vet[i] == '4') {
        res.push_back('2');
        res.push_back('2');
        res.push_back('3');
      } else if (vet[i] == '6') {
        res.push_back('3');
        res.push_back('5');
      } else if (vet[i] == '8') {
        res.push_back('2');
        res.push_back('2');
        res.push_back('2');
        res.push_back('7');
      } else if (vet[i] == '9') {
        res.push_back('3');
        res.push_back('3');
        res.push_back('2');
        res.push_back('7');
      } else {
        res.push_back(vet[i]);
      }
    }
  }
  sort(res.begin(), res.end());
  for (i = res.size() - 1; i >= 0; i--) {
    printf("%c", res[i]);
  }
  printf("\n");
  return 0;
}

### Prompt

Generate 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;
  char vet[20];
  vector<char> res;
  scanf("%d", &n);
  scanf("%s", vet);
  int i;
  for (i = 0; i < n; i++) {
    if (vet[i] > '1') {
      if (vet[i] == '4') {
        res.push_back('2');
        res.push_back('2');
        res.push_back('3');
      } else if (vet[i] == '6') {
        res.push_back('3');
        res.push_back('5');
      } else if (vet[i] == '8') {
        res.push_back('2');
        res.push_back('2');
        res.push_back('2');
        res.push_back('7');
      } else if (vet[i] == '9') {
        res.push_back('3');
        res.push_back('3');
        res.push_back('2');
        res.push_back('7');
      } else {
        res.push_back(vet[i]);
      }
    }
  }
  sort(res.begin(), res.end());
  for (i = res.size() - 1; i >= 0; i--) {
    printf("%c", res[i]);
  }
  printf("\n");
  return 0;
}

```