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>
#include <bits/stdc++.h>
using namespace std;
string s;
int n;
vector<int> r;
int main() {
  ios ::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> s;
  for (int i = 0; i < s.size(); i++) {
    int tmp = s[i] - '0';
    if (tmp == 0 || tmp == 1) continue;
    if (tmp == 4) {
      r.push_back(2);
      r.push_back(2);
      r.push_back(3);
    } else if (tmp == 6) {
      r.push_back(5);
      r.push_back(3);
    } else if (tmp == 8) {
      r.push_back(2);
      r.push_back(2);
      r.push_back(2);
      r.push_back(7);
    } else if (tmp == 9) {
      r.push_back(7);
      r.push_back(3);
      r.push_back(3);
      r.push_back(2);
    } else
      r.push_back(tmp);
  }
  sort(r.begin(), r.end());
  reverse(r.begin(), r.end());
  for (int i = 0; i < r.size(); i++) cout << r[i];
  cout << endl;
  return 0;
}
