#include <bits/stdc++.h>
using namespace std;
const int INF = numeric_limits<int>::max();
const long long LLINF = numeric_limits<long long>::max();
const unsigned long long ULLINF = numeric_limits<unsigned long long>::max();
string s[] = {"", "", "2", "3", "322", "5", "35", "7", "7222", "7332"};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  string a;
  cin >> a;
  string x = "";
  for (char f : a) x += s[f - '0'];
  sort((x).begin(), (x).end());
  reverse((x).begin(), (x).end());
  cout << x;
  return 0;
}

### Prompt

Your task is to create 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;
const int INF = numeric_limits<int>::max();
const long long LLINF = numeric_limits<long long>::max();
const unsigned long long ULLINF = numeric_limits<unsigned long long>::max();
string s[] = {"", "", "2", "3", "322", "5", "35", "7", "7222", "7332"};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  string a;
  cin >> a;
  string x = "";
  for (char f : a) x += s[f - '0'];
  sort((x).begin(), (x).end());
  reverse((x).begin(), (x).end());
  cout << x;
  return 0;
}

```