#include <bits/stdc++.h>
using namespace std;
int main() {
  string a[10];
  a[1] = "1";
  a[2] = "2";
  a[3] = "3";
  a[4] = "322";
  a[5] = "5";
  a[6] = "35";
  a[7] = "7";
  a[8] = "7222";
  a[9] = "7332";
  long long int n;
  scanf("%I64d", &n);
  getchar();
  string s;
  cin >> s;
  vector<string> v;
  for (long long int i = 0; i < n; i++) {
    if (s[i] > '1' && s[i] <= '9') {
      long long int temp = s[i] - '0';
      v.push_back(a[temp]);
    }
  }
  string q;
  vector<string>::iterator itr;
  for (itr = v.begin(); itr != v.end(); itr++) {
    q += *itr;
  }
  sort(q.begin(), q.end());
  reverse(q.begin(), q.end());
  cout << q << endl;
}

### 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 main() {
  string a[10];
  a[1] = "1";
  a[2] = "2";
  a[3] = "3";
  a[4] = "322";
  a[5] = "5";
  a[6] = "35";
  a[7] = "7";
  a[8] = "7222";
  a[9] = "7332";
  long long int n;
  scanf("%I64d", &n);
  getchar();
  string s;
  cin >> s;
  vector<string> v;
  for (long long int i = 0; i < n; i++) {
    if (s[i] > '1' && s[i] <= '9') {
      long long int temp = s[i] - '0';
      v.push_back(a[temp]);
    }
  }
  string q;
  vector<string>::iterator itr;
  for (itr = v.begin(); itr != v.end(); itr++) {
    q += *itr;
  }
  sort(q.begin(), q.end());
  reverse(q.begin(), q.end());
  cout << q << endl;
}

```