#include <bits/stdc++.h>
using namespace std;
int n;
char s[20];
int ans[4 * 20];
int main() {
  scanf("%d", &n);
  scanf("%s", &s);
  int m = 0;
  for (int i = 0; i < n; ++i) {
    if (s[i] == '0' || s[i] == '1')
      ;
    else if (s[i] == '2')
      ans[++m] = 2;
    else if (s[i] == '3')
      ans[++m] = 3;
    else if (s[i] == '4') {
      ans[++m] = 3;
      ans[++m] = 2;
      ans[++m] = 2;
    } else if (s[i] == '5')
      ans[++m] = 5;
    else if (s[i] == '6') {
      ans[++m] = 5;
      ans[++m] = 3;
    } else if (s[i] == '7')
      ans[++m] = 7;
    else if (s[i] == '8') {
      ans[++m] = 7;
      ans[++m] = 2;
      ans[++m] = 2;
      ans[++m] = 2;
    } else if (s[i] == '9') {
      ans[++m] = 7;
      ans[++m] = 3;
      ans[++m] = 3;
      ans[++m] = 2;
    }
  }
  sort(ans + 1, ans + m + 1);
  for (int i = m; i > 0; --i) printf("%d", ans[i]);
  puts("");
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 n;
char s[20];
int ans[4 * 20];
int main() {
  scanf("%d", &n);
  scanf("%s", &s);
  int m = 0;
  for (int i = 0; i < n; ++i) {
    if (s[i] == '0' || s[i] == '1')
      ;
    else if (s[i] == '2')
      ans[++m] = 2;
    else if (s[i] == '3')
      ans[++m] = 3;
    else if (s[i] == '4') {
      ans[++m] = 3;
      ans[++m] = 2;
      ans[++m] = 2;
    } else if (s[i] == '5')
      ans[++m] = 5;
    else if (s[i] == '6') {
      ans[++m] = 5;
      ans[++m] = 3;
    } else if (s[i] == '7')
      ans[++m] = 7;
    else if (s[i] == '8') {
      ans[++m] = 7;
      ans[++m] = 2;
      ans[++m] = 2;
      ans[++m] = 2;
    } else if (s[i] == '9') {
      ans[++m] = 7;
      ans[++m] = 3;
      ans[++m] = 3;
      ans[++m] = 2;
    }
  }
  sort(ans + 1, ans + m + 1);
  for (int i = m; i > 0; --i) printf("%d", ans[i]);
  puts("");
  return 0;
}

```