#include <bits/stdc++.h>
using namespace std;
int main() {
  int i, j, k, n;
  scanf("%d", &n);
  char s[n + 9];
  scanf("%s", s);
  char ans[n + 123];
  k = 0;
  for (i = 0; i < n; i++) {
    j = s[i] - '2';
    if (j < 0) continue;
    if (j < 2 || j == 3 || j == 5)
      ans[k++] = j + '2';
    else if (j == 2) {
      ans[k++] = j + '0';
      ans[k++] = j + '0';
      ans[k++] = j + '1';
    } else if (j == 4) {
      ans[k++] = j + '0' - 1;
      ans[k++] = j + '1';
    } else if (j == 6) {
      ans[k++] = j + '1';
      ans[k++] = j + '0' - 4;
      ans[k++] = j + '0' - 4;
      ans[k++] = j + '0' - 4;
    } else {
      ans[k++] = j + '0';
      ans[k++] = j + '0' - 4;
      ans[k++] = j + '0' - 4;
      ans[k++] = j + '0' - 5;
    }
  }
  sort(ans, ans + k);
  reverse(ans, ans + k);
  ans[k] = '\0';
  printf("%s\n", ans);
}

### Prompt

Create a solution in Cpp for 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 i, j, k, n;
  scanf("%d", &n);
  char s[n + 9];
  scanf("%s", s);
  char ans[n + 123];
  k = 0;
  for (i = 0; i < n; i++) {
    j = s[i] - '2';
    if (j < 0) continue;
    if (j < 2 || j == 3 || j == 5)
      ans[k++] = j + '2';
    else if (j == 2) {
      ans[k++] = j + '0';
      ans[k++] = j + '0';
      ans[k++] = j + '1';
    } else if (j == 4) {
      ans[k++] = j + '0' - 1;
      ans[k++] = j + '1';
    } else if (j == 6) {
      ans[k++] = j + '1';
      ans[k++] = j + '0' - 4;
      ans[k++] = j + '0' - 4;
      ans[k++] = j + '0' - 4;
    } else {
      ans[k++] = j + '0';
      ans[k++] = j + '0' - 4;
      ans[k++] = j + '0' - 4;
      ans[k++] = j + '0' - 5;
    }
  }
  sort(ans, ans + k);
  reverse(ans, ans + k);
  ans[k] = '\0';
  printf("%s\n", ans);
}

```