#include <bits/stdc++.h>
int main() {
  int n, i, ar[4];
  char s[20], ans[10000];
  scanf("%d %s", &n, s);
  ar[0] = 0;
  ar[1] = 0;
  ar[2] = 0;
  ar[3] = 0;
  for (i = 0; i < n; i++) {
    switch (s[i]) {
      case '9':
        ar[1] += 2;
      case '8':
        ar[0] += 3;
      case '7':
        ar[3]++;
      case '6':
        ar[0]++;
        ar[1]++;
      case '5':
        ar[2]++;
      case '4':
        ar[0] += 2;
      case '3':
        ar[1]++;
      case '2':
        ar[0]++;
      case '1':
        break;
      case '0':
        break;
    }
  }
  i = 0;
  while (ar[3] > 0) {
    ans[i] = '7';
    ar[3]--;
    ar[2]--;
    ar[1] -= 2;
    ar[0] -= 4;
    i++;
  }
  while (ar[2] > 0) {
    ans[i] = '5';
    ar[2]--;
    ar[1]--;
    ar[0] -= 3;
    i++;
  }
  while (ar[1] > 0) {
    ans[i] = '3';
    ar[1]--;
    ar[0]--;
    i++;
  }
  while (ar[0] > 0) {
    ans[i] = '2';
    ar[0]--;
    i++;
  }
  ans[i] = '\0';
  printf("%s\n", ans);
  return 0;
}

### Prompt

Please create a solution in Cpp 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() {
  int n, i, ar[4];
  char s[20], ans[10000];
  scanf("%d %s", &n, s);
  ar[0] = 0;
  ar[1] = 0;
  ar[2] = 0;
  ar[3] = 0;
  for (i = 0; i < n; i++) {
    switch (s[i]) {
      case '9':
        ar[1] += 2;
      case '8':
        ar[0] += 3;
      case '7':
        ar[3]++;
      case '6':
        ar[0]++;
        ar[1]++;
      case '5':
        ar[2]++;
      case '4':
        ar[0] += 2;
      case '3':
        ar[1]++;
      case '2':
        ar[0]++;
      case '1':
        break;
      case '0':
        break;
    }
  }
  i = 0;
  while (ar[3] > 0) {
    ans[i] = '7';
    ar[3]--;
    ar[2]--;
    ar[1] -= 2;
    ar[0] -= 4;
    i++;
  }
  while (ar[2] > 0) {
    ans[i] = '5';
    ar[2]--;
    ar[1]--;
    ar[0] -= 3;
    i++;
  }
  while (ar[1] > 0) {
    ans[i] = '3';
    ar[1]--;
    ar[0]--;
    i++;
  }
  while (ar[0] > 0) {
    ans[i] = '2';
    ar[0]--;
    i++;
  }
  ans[i] = '\0';
  printf("%s\n", ans);
  return 0;
}

```