#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, count[9] = {0, 0, 0, 0, 0, 0, 0, 0, 0};
  char s;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("\n%c", &s);
    s -= '0';
    if (s == 2 || s == 3 || s == 5 || s == 7)
      count[s - 1]++;
    else if (s == 4) {
      count[2]++;
      count[1] += 2;
    } else if (s == 6) {
      count[4]++;
      count[2]++;
    } else if (s == 8) {
      count[6]++;
      count[1] += 3;
    } else if (s == 9) {
      count[6]++;
      count[2] += 2;
      count[1]++;
    }
  }
  for (int i = 8; i >= 0; i--) {
    while (count[i] > 0) {
      printf("%d", i + 1);
      count[i]--;
    }
  }
  printf("\n");
  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>
using namespace std;
int main() {
  int n, count[9] = {0, 0, 0, 0, 0, 0, 0, 0, 0};
  char s;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("\n%c", &s);
    s -= '0';
    if (s == 2 || s == 3 || s == 5 || s == 7)
      count[s - 1]++;
    else if (s == 4) {
      count[2]++;
      count[1] += 2;
    } else if (s == 6) {
      count[4]++;
      count[2]++;
    } else if (s == 8) {
      count[6]++;
      count[1] += 3;
    } else if (s == 9) {
      count[6]++;
      count[2] += 2;
      count[1]++;
    }
  }
  for (int i = 8; i >= 0; i--) {
    while (count[i] > 0) {
      printf("%d", i + 1);
      count[i]--;
    }
  }
  printf("\n");
  return 0;
}

```