#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  string str;
  cin >> n >> str;
  int cnt[10];
  for (int i = 0; i < 10; i++) cnt[i] = 0;
  for (int i = 0; i < n; i++) switch (str[i]) {
      case '2':
        ++cnt[2];
        break;
      case '3':
        ++cnt[3];
        break;
      case '4':
        ++cnt[3];
        cnt[2] += 2;
        break;
      case '5':
        ++cnt[5];
        break;
      case '6':
        ++cnt[5];
        ++cnt[3];
        break;
      case '7':
        ++cnt[7];
        break;
      case '8':
        ++cnt[7];
        cnt[2] += 3;
        break;
      case '9':
        ++cnt[7];
        cnt[3] += 2;
        ++cnt[2];
        break;
    }
  for (int i = 9; i > 0; i--)
    for (int j = 0; j < cnt[i]; j++) cout << i;
  cout << endl;
}

### 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>
using namespace std;
int main() {
  int n;
  string str;
  cin >> n >> str;
  int cnt[10];
  for (int i = 0; i < 10; i++) cnt[i] = 0;
  for (int i = 0; i < n; i++) switch (str[i]) {
      case '2':
        ++cnt[2];
        break;
      case '3':
        ++cnt[3];
        break;
      case '4':
        ++cnt[3];
        cnt[2] += 2;
        break;
      case '5':
        ++cnt[5];
        break;
      case '6':
        ++cnt[5];
        ++cnt[3];
        break;
      case '7':
        ++cnt[7];
        break;
      case '8':
        ++cnt[7];
        cnt[2] += 3;
        break;
      case '9':
        ++cnt[7];
        cnt[3] += 2;
        ++cnt[2];
        break;
    }
  for (int i = 9; i > 0; i--)
    for (int j = 0; j < cnt[i]; j++) cout << i;
  cout << endl;
}

```