#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, i, c;
  multiset<int> a[10];
  vector<int> b;
  multiset<int>::iterator it;
  a[2].insert(2);
  a[3].insert(3);
  a[4].insert(3);
  a[4].insert(2);
  a[4].insert(2);
  a[5].insert(5);
  a[6].insert(5);
  a[6].insert(3);
  a[7].insert(7);
  a[8].insert(7);
  a[8].insert(2);
  a[8].insert(2);
  a[8].insert(2);
  a[9].insert(7);
  a[9].insert(2);
  a[9].insert(3);
  a[9].insert(3);
  cin >> n;
  string s;
  cin >> s;
  for (i = 0; i < s.length(); i++) {
    c = s[i] - '0';
    if (c == '0' || c == '1') continue;
    for (it = a[c].begin(); it != a[c].end(); it++) {
      b.push_back(*it);
    }
  }
  sort(b.begin(), b.end(), greater<__typeof(b[0])>());
  for (i = 0; i < b.size(); i++) cout << b[i];
  cout << endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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, i, c;
  multiset<int> a[10];
  vector<int> b;
  multiset<int>::iterator it;
  a[2].insert(2);
  a[3].insert(3);
  a[4].insert(3);
  a[4].insert(2);
  a[4].insert(2);
  a[5].insert(5);
  a[6].insert(5);
  a[6].insert(3);
  a[7].insert(7);
  a[8].insert(7);
  a[8].insert(2);
  a[8].insert(2);
  a[8].insert(2);
  a[9].insert(7);
  a[9].insert(2);
  a[9].insert(3);
  a[9].insert(3);
  cin >> n;
  string s;
  cin >> s;
  for (i = 0; i < s.length(); i++) {
    c = s[i] - '0';
    if (c == '0' || c == '1') continue;
    for (it = a[c].begin(); it != a[c].end(); it++) {
      b.push_back(*it);
    }
  }
  sort(b.begin(), b.end(), greater<__typeof(b[0])>());
  for (i = 0; i < b.size(); i++) cout << b[i];
  cout << endl;
  return 0;
}

```