#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, i;
  cin >> n;
  string s;
  int arr[10];
  cin >> s;
  for (i = 0; i < 10; i++) {
    arr[i] = 0;
  }
  for (i = 0; i < n; i++) {
    arr[s[i] - 48]++;
  }
  while (arr[4] > 0) {
    arr[4]--;
    arr[3]++;
    arr[2] += 2;
  }
  while (arr[6] > 0) {
    arr[6]--;
    arr[5]++;
    arr[3]++;
  }
  while (arr[8] > 0) {
    arr[8]--;
    arr[7]++;
    arr[2] += 3;
  }
  while (arr[9] > 0) {
    arr[9]--;
    arr[7]++;
    arr[3] += 2;
    arr[2]++;
  }
  for (i = 9; i > 1; i--) {
    while (arr[i] > 0) {
      cout << i;
      arr[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 main() {
  int n, i;
  cin >> n;
  string s;
  int arr[10];
  cin >> s;
  for (i = 0; i < 10; i++) {
    arr[i] = 0;
  }
  for (i = 0; i < n; i++) {
    arr[s[i] - 48]++;
  }
  while (arr[4] > 0) {
    arr[4]--;
    arr[3]++;
    arr[2] += 2;
  }
  while (arr[6] > 0) {
    arr[6]--;
    arr[5]++;
    arr[3]++;
  }
  while (arr[8] > 0) {
    arr[8]--;
    arr[7]++;
    arr[2] += 3;
  }
  while (arr[9] > 0) {
    arr[9]--;
    arr[7]++;
    arr[3] += 2;
    arr[2]++;
  }
  for (i = 9; i > 1; i--) {
    while (arr[i] > 0) {
      cout << i;
      arr[i]--;
    }
  }
}

```