#include <bits/stdc++.h>
using namespace std;
int bp(int a, int n) {
  int res = 1;
  while (n) {
    if (n & 1) res *= a;
    a *= a;
    n >>= 1;
  }
  return res;
}
void solve() {
  int n;
  cin >> n;
  string k;
  cin >> k;
  map<char, string> t = {
      {'0', "\0"}, {'1', "\0"}, {'2', "2"}, {'3', "3"},    {'4', "322"},
      {'5', "5"},  {'6', "53"}, {'7', "7"}, {'8', "7222"}, {'9', "7332"},
  };
  int ind = 0;
  while (ind < (int)(k).size()) {
    string r = t[k[ind]];
    k.erase(k.begin() + ind, k.begin() + ind + 1);
    k.insert(ind, r);
    ind += ((int)(r).size() > 0 ? (int)(r).size() : 0);
  }
  sort((k).begin(), (k).end(), greater<char>());
  cout << k;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) solve();
}

### Prompt

Your task is to create a cpp solution to the following problem:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int bp(int a, int n) {
  int res = 1;
  while (n) {
    if (n & 1) res *= a;
    a *= a;
    n >>= 1;
  }
  return res;
}
void solve() {
  int n;
  cin >> n;
  string k;
  cin >> k;
  map<char, string> t = {
      {'0', "\0"}, {'1', "\0"}, {'2', "2"}, {'3', "3"},    {'4', "322"},
      {'5', "5"},  {'6', "53"}, {'7', "7"}, {'8', "7222"}, {'9', "7332"},
  };
  int ind = 0;
  while (ind < (int)(k).size()) {
    string r = t[k[ind]];
    k.erase(k.begin() + ind, k.begin() + ind + 1);
    k.insert(ind, r);
    ind += ((int)(r).size() > 0 ? (int)(r).size() : 0);
  }
  sort((k).begin(), (k).end(), greater<char>());
  cout << k;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) solve();
}

```