#include <bits/stdc++.h>
using namespace std;
int b[8];
string str;
int main() {
  int n, i;
  cin >> n >> str;
  for (i = 0; i <= (n - 1); i++) switch (str[i]) {
      case '2':
        b[2]++;
        break;
      case '3':
        b[3]++;
        break;
      case '4':
        b[2] += 2, b[3]++;
        break;
      case '5':
        b[5]++;
        break;
      case '6':
        b[5]++, b[3]++;
        break;
      case '7':
        b[7]++;
        break;
      case '8':
        b[7]++, b[2] += 3;
        break;
      case '9': {
        b[7]++;
        b[2]++;
        b[3] += 2;
      }
    }
  while (b[7]--) putchar('7');
  while (b[5]--) putchar('5');
  while (b[3]--) putchar('3');
  while (b[2]--) putchar('2');
  return 0;
}

### Prompt

Your task is to create 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>
using namespace std;
int b[8];
string str;
int main() {
  int n, i;
  cin >> n >> str;
  for (i = 0; i <= (n - 1); i++) switch (str[i]) {
      case '2':
        b[2]++;
        break;
      case '3':
        b[3]++;
        break;
      case '4':
        b[2] += 2, b[3]++;
        break;
      case '5':
        b[5]++;
        break;
      case '6':
        b[5]++, b[3]++;
        break;
      case '7':
        b[7]++;
        break;
      case '8':
        b[7]++, b[2] += 3;
        break;
      case '9': {
        b[7]++;
        b[2]++;
        b[3] += 2;
      }
    }
  while (b[7]--) putchar('7');
  while (b[5]--) putchar('5');
  while (b[3]--) putchar('3');
  while (b[2]--) putchar('2');
  return 0;
}

```