#include <bits/stdc++.h>
using namespace std;
int res[1000];
int main() {
  int n;
  long long s;
  cin >> n;
  cin >> s;
  int k = 0;
  for (int i = 1; i <= n; i++) {
    int tmp = s % 10;
    s /= 10;
    if (tmp == 6) {
      res[++k] = 5;
      res[++k] = 3;
    } else if (tmp == 4) {
      res[++k] = 3;
      res[++k] = 2;
      res[++k] = 2;
    } else if (tmp == 8) {
      for (int j = 1; j <= 3; j++) res[++k] = 2;
      res[++k] = 7;
    } else if (tmp == 9) {
      res[++k] = 3;
      res[++k] = 3;
      res[++k] = 2;
      res[++k] = 7;
    } else if (tmp != 1 && tmp != 0)
      res[++k] = tmp;
  }
  sort(res + 1, res + k + 1);
  for (int i = k; i >= 1; i--) cout << res[i];
}

### Prompt

Your challenge is to write 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>
using namespace std;
int res[1000];
int main() {
  int n;
  long long s;
  cin >> n;
  cin >> s;
  int k = 0;
  for (int i = 1; i <= n; i++) {
    int tmp = s % 10;
    s /= 10;
    if (tmp == 6) {
      res[++k] = 5;
      res[++k] = 3;
    } else if (tmp == 4) {
      res[++k] = 3;
      res[++k] = 2;
      res[++k] = 2;
    } else if (tmp == 8) {
      for (int j = 1; j <= 3; j++) res[++k] = 2;
      res[++k] = 7;
    } else if (tmp == 9) {
      res[++k] = 3;
      res[++k] = 3;
      res[++k] = 2;
      res[++k] = 7;
    } else if (tmp != 1 && tmp != 0)
      res[++k] = tmp;
  }
  sort(res + 1, res + k + 1);
  for (int i = k; i >= 1; i--) cout << res[i];
}

```