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 int n, m, idx, counte;
string s;
char c;
vector<long long int> v;
inline void solve(void) {
  cin >> n;
  cin >> s;
  for (int i = (0); i < (n); i++) {
    long long int x = s[i] - '0';
    if (x != 0 && x != 1) {
      v.push_back(x);
    }
  }
  vector<long long int> ans;
  long long int len = v.size();
  for (int i = (0); i < (len); i++) {
    if (v[i] == 4) {
      ans.push_back(3);
      ans.push_back(2);
      ans.push_back(2);
    } else if (v[i] == 6) {
      ans.push_back(5);
      ans.push_back(3);
    } else if (v[i] == 8) {
      ans.push_back(7);
      ans.push_back(2);
      ans.push_back(2);
      ans.push_back(2);
    } else if (v[i] == 9) {
      ans.push_back(7);
      ans.push_back(3);
      ans.push_back(3);
      ans.push_back(2);
    } else
      ans.push_back(v[i]);
  }
  sort(ans.rbegin(), ans.rend());
  len = ans.size();
  for (int i = (0); i < (len); i++) cout << ans[i];
}
inline void recurse(int n) {}
int main() {
  long long int test = 1;
  while (test--) {
    solve();
  }
  return 0;
}
