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;
int n;
long long x;
int d[20];
vector<int> sol;
int main() {
  cin >> n;
  cin >> x;
  n = 0;
  while (x) {
    ++n;
    d[n] = x % 10;
    x /= 10;
  }
  for (int i = 1; i <= n; ++i) {
    if (d[i] == 4) {
      sol.push_back(2);
      sol.push_back(2);
      sol.push_back(3);
    } else if (d[i] == 6) {
      sol.push_back(3);
      sol.push_back(5);
    } else if (d[i] == 8) {
      sol.push_back(2);
      sol.push_back(2);
      sol.push_back(2);
      sol.push_back(7);
    } else if (d[i] == 9) {
      sol.push_back(3);
      sol.push_back(3);
      sol.push_back(2);
      sol.push_back(7);
    } else
      sol.push_back(d[i]);
  }
  sort(sol.begin(), sol.end());
  reverse(sol.begin(), sol.end());
  for (int i = 0; i < (int)sol.size(); ++i) {
    if (sol[i] > 1) cout << sol[i];
  }
  cout << "\n";
  return 0;
}
