#include <bits/stdc++.h>
int temp[10], i, x, n;
char arr[100];
void ekle(char x) {
  switch (x) {
    case '4':
      temp[3]++, temp[2] += 2;
      break;
    case '6':
      temp[5]++, temp[3]++;
      break;
    case '8':
      temp[7]++, temp[2] += 3;
      break;
    case '9':
      temp[7]++, temp[3] += 2, temp[2]++;
      break;
    default:
      temp[x - '0']++;
  }
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf(" %c", &arr[i]), ekle(arr[i]);
  for (i = 9; i >= 2; i--)
    while (temp[i]--) printf("%d", i);
  return 0;
}

### 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>
int temp[10], i, x, n;
char arr[100];
void ekle(char x) {
  switch (x) {
    case '4':
      temp[3]++, temp[2] += 2;
      break;
    case '6':
      temp[5]++, temp[3]++;
      break;
    case '8':
      temp[7]++, temp[2] += 3;
      break;
    case '9':
      temp[7]++, temp[3] += 2, temp[2]++;
      break;
    default:
      temp[x - '0']++;
  }
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf(" %c", &arr[i]), ekle(arr[i]);
  for (i = 9; i >= 2; i--)
    while (temp[i]--) printf("%d", i);
  return 0;
}

```