#include <bits/stdc++.h>
using namespace std;
int result[10];
int main() {
  long long int count = 0, num;
  cin >> count >> num;
  while (num > 0) {
    int temp = num % 10;
    if (temp == 0 || temp == 1)
      result[1]++;
    else if (temp == 2)
      result[2]++;
    else if (temp == 3)
      result[3]++;
    else if (temp == 4) {
      result[3]++;
      result[2] += 2;
    } else if (temp == 5)
      result[5]++;
    else if (temp == 6) {
      result[3]++;
      result[5]++;
    } else if (temp == 7)
      result[7]++;
    else if (temp == 8) {
      result[7]++;
      result[2] += 3;
    } else {
      result[7]++;
      result[3] += 2;
      result[2]++;
    }
    num = num / 10;
  }
  for (int i = 9; i > 1; i--)
    while (result[i]--) cout << i;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 result[10];
int main() {
  long long int count = 0, num;
  cin >> count >> num;
  while (num > 0) {
    int temp = num % 10;
    if (temp == 0 || temp == 1)
      result[1]++;
    else if (temp == 2)
      result[2]++;
    else if (temp == 3)
      result[3]++;
    else if (temp == 4) {
      result[3]++;
      result[2] += 2;
    } else if (temp == 5)
      result[5]++;
    else if (temp == 6) {
      result[3]++;
      result[5]++;
    } else if (temp == 7)
      result[7]++;
    else if (temp == 8) {
      result[7]++;
      result[2] += 3;
    } else {
      result[7]++;
      result[3] += 2;
      result[2]++;
    }
    num = num / 10;
  }
  for (int i = 9; i > 1; i--)
    while (result[i]--) cout << i;
}

```