#include <bits/stdc++.h>
using namespace std;
const long long oo = 0x3f3f3f3f3f3f3f3f;
const double EPS = 1e-7;
const int MAXN = (int)1e5 + 10;
char S[100];
vector<char> ans;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  cin >> S;
  for (int i = 0; i < n; ++i) {
    int val = S[i] - '0';
    if (val <= 1) continue;
    if (val == 2) ans.push_back('2');
    if (val == 3) ans.push_back('3');
    if (val == 4) {
      ans.push_back('3');
      ans.push_back('2');
      ans.push_back('2');
    }
    if (val == 5) ans.push_back('5');
    if (val == 6) {
      ans.push_back('5');
      ans.push_back('3');
    }
    if (val == 7) ans.push_back('7');
    if (val == 8) {
      ans.push_back('7');
      ans.push_back('2');
      ans.push_back('2');
      ans.push_back('2');
    }
    if (val == 9) {
      ans.push_back('7');
      ans.push_back('3');
      ans.push_back('3');
      ans.push_back('2');
    }
  }
  sort(ans.begin(), ans.end());
  for (int i = ((int)((ans).size())) - 1; i >= 0; --i) cout << ans[i];
  cout << endl;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
const long long oo = 0x3f3f3f3f3f3f3f3f;
const double EPS = 1e-7;
const int MAXN = (int)1e5 + 10;
char S[100];
vector<char> ans;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  cin >> S;
  for (int i = 0; i < n; ++i) {
    int val = S[i] - '0';
    if (val <= 1) continue;
    if (val == 2) ans.push_back('2');
    if (val == 3) ans.push_back('3');
    if (val == 4) {
      ans.push_back('3');
      ans.push_back('2');
      ans.push_back('2');
    }
    if (val == 5) ans.push_back('5');
    if (val == 6) {
      ans.push_back('5');
      ans.push_back('3');
    }
    if (val == 7) ans.push_back('7');
    if (val == 8) {
      ans.push_back('7');
      ans.push_back('2');
      ans.push_back('2');
      ans.push_back('2');
    }
    if (val == 9) {
      ans.push_back('7');
      ans.push_back('3');
      ans.push_back('3');
      ans.push_back('2');
    }
  }
  sort(ans.begin(), ans.end());
  for (int i = ((int)((ans).size())) - 1; i >= 0; --i) cout << ans[i];
  cout << endl;
}

```