#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T ncr(T n, T r) {
  return n * (n - 1) / r;
}
string originate[] = {"", "", "2", "3", "322", "5", "53", "7", "7222", "7332"};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  string s, ans = "";
  cin >> n >> s;
  for (int i = 0; s[i]; i++) ans += originate[s[i] - '0'];
  sort(ans.begin(), ans.end());
  reverse(ans.begin(), ans.end());
  cout << ans << "\n";
  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;
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T ncr(T n, T r) {
  return n * (n - 1) / r;
}
string originate[] = {"", "", "2", "3", "322", "5", "53", "7", "7222", "7332"};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  string s, ans = "";
  cin >> n >> s;
  for (int i = 0; s[i]; i++) ans += originate[s[i] - '0'];
  sort(ans.begin(), ans.end());
  reverse(ans.begin(), ans.end());
  cout << ans << "\n";
  return 0;
}

```