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;
template <int POS, class TUPLE>
void deploy(std::ostream &os, const TUPLE &tuple) {}
template <int POS, class TUPLE, class H, class... Ts>
void deploy(std::ostream &os, const TUPLE &t) {
  os << (POS == 0 ? "" : ", ") << get<POS>(t);
  deploy<POS + 1, TUPLE, Ts...>(os, t);
}
template <class T, class U>
std::ostream &operator<<(std::ostream &os, std::pair<T, U> &p) {
  os << "(" << p.first << ", " << p.second << ")";
  return os;
}
template <class T>
std::ostream &operator<<(std::ostream &os, std::vector<T> &v) {
  int remain = v.size();
  os << "{";
  for (auto e : v) os << e << (--remain == 0 ? "}" : ", ");
  return os;
}
template <class T>
std::ostream &operator<<(std::ostream &os, std::set<T> &v) {
  int remain = v.size();
  os << "{";
  for (auto e : v) os << e << (--remain == 0 ? "}" : ", ");
  return os;
}
template <class T, class K>
std::ostream &operator<<(std::ostream &os, std::map<T, K> &make_pair) {
  int remain = make_pair.size();
  os << "{";
  for (auto e : make_pair)
    os << "(" << e.first << " -> " << e.second << ")"
       << (--remain == 0 ? "}" : ", ");
  return os;
}
using ll = long long;
int N;
string S;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> N;
  cin >> S;
  {
    string newS;
    for (int i = 0; i < (((int)(S).size())); i++) {
      if (S[i] == '1' || S[i] == '0') {
        continue;
      }
      newS += S[i];
    }
    S = newS;
  }
  string ans;
  for (int i = 0; i < (((int)(S).size())); i++) {
    auto x = S[i] - '0';
    switch (x) {
      case 2:
        ans += '2';
        break;
      case 3:
        ans += '3';
        break;
      case 4:
        ans += "322";
        break;
      case 5:
        ans += '5';
        break;
      case 6:
        ans += "35";
        break;
      case 7:
        ans += '7';
        break;
      case 8:
        ans += "7222";
        break;
      case 9:
        ans += "7332";
        break;
    }
  }
  sort((ans).begin(), (ans).end(), greater<char>());
  cout << ans << "\n";
  return 0;
}
