#include <bits/stdc++.h>
int main(void) {
  char a[60], b[100];
  int i = 0, j = 0, n, J, t;
  scanf("%d", &n);
  getchar();
  gets(a);
  memset(b, 0, sizeof(b));
  for (i = 0; i < n; i++) {
    switch (a[i]) {
      case '0':
        break;
      case '1':
        break;
      case '2':
        b[j] = a[i];
        j++;
        break;
      case '3':
        b[j] = a[i];
        j++;
        break;
      case '4':
        b[j] = '3';
        b[j + 1] = '2';
        b[j + 2] = '2';
        j = j + 3;
        break;
      case '5':
        b[j] = a[i];
        j++;
        break;
      case '6':
        b[j] = '5';
        b[j + 1] = '3';
        j = j + 2;
        break;
      case '7':
        b[j] = a[i];
        j++;
        break;
      case '8':
        b[j] = '7';
        b[j + 1] = '2';
        b[j + 2] = '2';
        b[j + 3] = '2';
        j = j + 4;
        break;
      case '9':
        b[j] = '7';
        b[j + 1] = '3';
        b[j + 2] = '3';
        b[j + 3] = '2';
        j = j + 4;
        break;
    }
  }
  J = strlen(b);
  for (j = 0; j < J; j++) {
    for (i = 0; i < J - j; i++) {
      if (b[i] < b[i + 1]) {
        t = b[i];
        b[i] = b[i + 1];
        b[i + 1] = t;
      }
    }
  }
  printf("%s\n", b);
  return 0;
}

### Prompt

In CPP, your task is to solve 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(void) {
  char a[60], b[100];
  int i = 0, j = 0, n, J, t;
  scanf("%d", &n);
  getchar();
  gets(a);
  memset(b, 0, sizeof(b));
  for (i = 0; i < n; i++) {
    switch (a[i]) {
      case '0':
        break;
      case '1':
        break;
      case '2':
        b[j] = a[i];
        j++;
        break;
      case '3':
        b[j] = a[i];
        j++;
        break;
      case '4':
        b[j] = '3';
        b[j + 1] = '2';
        b[j + 2] = '2';
        j = j + 3;
        break;
      case '5':
        b[j] = a[i];
        j++;
        break;
      case '6':
        b[j] = '5';
        b[j + 1] = '3';
        j = j + 2;
        break;
      case '7':
        b[j] = a[i];
        j++;
        break;
      case '8':
        b[j] = '7';
        b[j + 1] = '2';
        b[j + 2] = '2';
        b[j + 3] = '2';
        j = j + 4;
        break;
      case '9':
        b[j] = '7';
        b[j + 1] = '3';
        b[j + 2] = '3';
        b[j + 3] = '2';
        j = j + 4;
        break;
    }
  }
  J = strlen(b);
  for (j = 0; j < J; j++) {
    for (i = 0; i < J - j; i++) {
      if (b[i] < b[i + 1]) {
        t = b[i];
        b[i] = b[i + 1];
        b[i + 1] = t;
      }
    }
  }
  printf("%s\n", b);
  return 0;
}

```