#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  char c[20];
  cin >> n;
  cin >> c;
  vector<int> v, r;
  for (int i = 0; i < strlen(c); i++)
    if (c[i] > '1') v.push_back((int)c[i] - '0');
  for (int i = 0; i < (v.size()); i++) {
    if (v[i] == 9) {
      r.push_back(7);
      r.push_back(3);
      r.push_back(3);
      r.push_back(2);
    } else if (v[i] == 8) {
      r.push_back(7);
      r.push_back(2);
      r.push_back(2);
      r.push_back(2);
    } else if (v[i] == 6) {
      r.push_back(5);
      r.push_back(3);
    } else if (v[i] == 4) {
      r.push_back(3);
      r.push_back(2);
      r.push_back(2);
    } else {
      r.push_back(v[i]);
    }
  }
  sort(r.begin(), r.end());
  for (int i = r.size() - 1; i >= 0; i--) cout << r[i];
  return 0;
}

### Prompt

Please formulate 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 main() {
  int n;
  char c[20];
  cin >> n;
  cin >> c;
  vector<int> v, r;
  for (int i = 0; i < strlen(c); i++)
    if (c[i] > '1') v.push_back((int)c[i] - '0');
  for (int i = 0; i < (v.size()); i++) {
    if (v[i] == 9) {
      r.push_back(7);
      r.push_back(3);
      r.push_back(3);
      r.push_back(2);
    } else if (v[i] == 8) {
      r.push_back(7);
      r.push_back(2);
      r.push_back(2);
      r.push_back(2);
    } else if (v[i] == 6) {
      r.push_back(5);
      r.push_back(3);
    } else if (v[i] == 4) {
      r.push_back(3);
      r.push_back(2);
      r.push_back(2);
    } else {
      r.push_back(v[i]);
    }
  }
  sort(r.begin(), r.end());
  for (int i = r.size() - 1; i >= 0; i--) cout << r[i];
  return 0;
}

```