#include <bits/stdc++.h>
int n, a[10], i, j;
char c;
int main() {
  std::cin >> n;
  for (i = 0; i < n; i++) {
    std::cin >> c;
    c -= 48;
    switch (c) {
      case 2:
      case 3:
      case 5:
      case 7: {
        a[c]++;
        break;
      }
      case 4: {
        a[3]++;
        a[2] += 2;
        break;
      }
      case 6: {
        a[5]++;
        a[3]++;
        break;
      }
      case 8: {
        a[7]++;
        a[2] += 3;
        break;
      }
      case 9: {
        a[7]++;
        a[3] += 2;
        a[2]++;
        break;
      }
    }
  }
  for (i = 9; i > 1; i--) {
    for (j = 0; j < a[i]; j++) std::cout << i;
  }
  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>
int n, a[10], i, j;
char c;
int main() {
  std::cin >> n;
  for (i = 0; i < n; i++) {
    std::cin >> c;
    c -= 48;
    switch (c) {
      case 2:
      case 3:
      case 5:
      case 7: {
        a[c]++;
        break;
      }
      case 4: {
        a[3]++;
        a[2] += 2;
        break;
      }
      case 6: {
        a[5]++;
        a[3]++;
        break;
      }
      case 8: {
        a[7]++;
        a[2] += 3;
        break;
      }
      case 9: {
        a[7]++;
        a[3] += 2;
        a[2]++;
        break;
      }
    }
  }
  for (i = 9; i > 1; i--) {
    for (j = 0; j < a[i]; j++) std::cout << i;
  }
  return 0;
}

```