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 gcd(long long int a, long long int b) {
  if (b == 0) return a;
  a %= b;
  return gcd(b, a);
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long int n;
  cin >> n;
  string s;
  cin >> s;
  vector<int> Fa(10, 0);
  for (long long int i = 0; i < n; i++) {
    if (s[i] == '2') Fa[2]++;
    if (s[i] == '3') Fa[3]++;
    if (s[i] == '4') Fa[2] += 2, Fa[3]++;
    if (s[i] == '5') Fa[5]++;
    if (s[i] == '6') Fa[3]++, Fa[5]++;
    if (s[i] == '7') Fa[7]++;
    if (s[i] == '8') Fa[2] += 3, Fa[7]++;
    if (s[i] == '9') Fa[3] += 2, Fa[2]++, Fa[7]++;
  }
  string ans = "";
  for (int i = 9; i >= 2; i--) {
    for (long long int j = 0; j < Fa[i]; j++) {
      ans = ans + (char)('0' + i);
    }
  }
  cout << ans << '\n';
  return 0;
}
