#include <bits/stdc++.h>
int main() {
  char data[10][10] = {"1", "1",  "2", "3",    "322",
                       "5", "53", "7", "7222", "7332"};
  char num[16];
  char ans[20];
  int n;
  int l, l1, l2, len;
  char tmp;
  while (2 == scanf("%d%s", &n, num)) {
    for (l = 0, len = 0; num[l]; l++) {
      if (num[l] > '1') {
        for (l1 = len, l2 = 0; data[num[l] - '0'][l2];
             ans[l1] = data[num[l] - '0'][l2], l2++, l1++)
          ;
        len = l1;
      }
    }
    for (l = 0; l < len; l++) {
      l2 = len - l;
      for (l1 = 1; l1 < l2; l1++) {
        if (ans[l1] > ans[l1 - 1]) {
          tmp = ans[l1];
          ans[l1] = ans[l1 - 1];
          ans[l1 - 1] = tmp;
        }
      }
    }
    ans[len] = '\0';
    puts(ans);
  }
  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>
int main() {
  char data[10][10] = {"1", "1",  "2", "3",    "322",
                       "5", "53", "7", "7222", "7332"};
  char num[16];
  char ans[20];
  int n;
  int l, l1, l2, len;
  char tmp;
  while (2 == scanf("%d%s", &n, num)) {
    for (l = 0, len = 0; num[l]; l++) {
      if (num[l] > '1') {
        for (l1 = len, l2 = 0; data[num[l] - '0'][l2];
             ans[l1] = data[num[l] - '0'][l2], l2++, l1++)
          ;
        len = l1;
      }
    }
    for (l = 0; l < len; l++) {
      l2 = len - l;
      for (l1 = 1; l1 < l2; l1++) {
        if (ans[l1] > ans[l1 - 1]) {
          tmp = ans[l1];
          ans[l1] = ans[l1 - 1];
          ans[l1 - 1] = tmp;
        }
      }
    }
    ans[len] = '\0';
    puts(ans);
  }
  return 0;
}

```