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;
long long fact[10];
int a[15];
vector<int> res;
int main() {
  fact[0] = 1;
  fact[1] = 1;
  for (int i = 2; i < 10; i++) {
    fact[i] = i * fact[i - 1];
  }
  int n;
  cin >> n;
  string s;
  cin >> s;
  int t = 0;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] >= '2' && s[i] <= '9') {
      a[t] = s[i] - '0';
      t++;
    }
  }
  for (int i = 0; i < t; i++) {
    if (a[i] == 4) {
      res.push_back(3);
      res.push_back(2);
      res.push_back(2);
    } else if (a[i] == 6) {
      res.push_back(5);
      res.push_back(3);
    } else if (a[i] == 8) {
      res.push_back(7);
      res.push_back(2);
      res.push_back(2);
      res.push_back(2);
    } else if (a[i] == 9) {
      res.push_back(7);
      res.push_back(2);
      res.push_back(3);
      res.push_back(3);
    } else
      res.push_back(a[i]);
  }
  sort(res.begin(), res.end());
  for (int i = res.size() - 1; i >= 0; i--) {
    cout << res[i];
  }
  cout << endl;
}
