#include <bits/stdc++.h>
using namespace std;
int main() {
  char num[100000];
  long long int n;
  cin >> n;
  cin >> num;
  long long int i, j, p, l, num1[100000], len = 0;
  char str[][100000] = {
      "2", "3", "322", "5", "53", "7", "7222", "7332",
  };
  for (i = 0; i < n; i++) {
    if (num[i] != '0' && num[i] != '1') {
      p = num[i] - 48 - 2;
      l = strlen(str[p]);
      for (j = 0; j < l; j++) {
        num1[len] = str[p][j] - 48;
        len++;
      }
    }
  }
  sort(num1, num1 + len);
  for (i = 0; i < len; i++) cout << num1[len - i - 1];
  return 0;
}

### Prompt

Your task is to create 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() {
  char num[100000];
  long long int n;
  cin >> n;
  cin >> num;
  long long int i, j, p, l, num1[100000], len = 0;
  char str[][100000] = {
      "2", "3", "322", "5", "53", "7", "7222", "7332",
  };
  for (i = 0; i < n; i++) {
    if (num[i] != '0' && num[i] != '1') {
      p = num[i] - 48 - 2;
      l = strlen(str[p]);
      for (j = 0; j < l; j++) {
        num1[len] = str[p][j] - 48;
        len++;
      }
    }
  }
  sort(num1, num1 + len);
  for (i = 0; i < len; i++) cout << num1[len - i - 1];
  return 0;
}

```