#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  long long no;
  cin >> no;
  list<int> prime_nos = {2, 3, 5, 7}, pn = prime_nos;
  map<int, int> factorial;
  while (no != 0) {
    long long n = no % 10;
    if (n == 2)
      factorial[2]++;
    else if (n == 3)
      factorial[3]++, factorial[2]++;
    else if (n == 4)
      factorial[3]++, factorial[2] += 3;
    else if (n == 5)
      factorial[5]++, factorial[3]++, factorial[2] += 3;
    else if (n == 6)
      factorial[5]++, factorial[3] += 2, factorial[2] += 4;
    else if (n == 7)
      factorial[7]++, factorial[5]++, factorial[3] += 2, factorial[2] += 4;
    else if (n == 8)
      factorial[7]++, factorial[5]++, factorial[3] += 2, factorial[2] += 7;
    else if (n == 9)
      factorial[7]++, factorial[5]++, factorial[3] += 4, factorial[2] += 7;
    no /= 10;
  }
  list<int> ans;
  for (auto it = factorial.rbegin(); it != factorial.rend(); it++) {
    if (it->first == 2) {
      ans.insert(ans.end(), it->second, 2);
    } else if (it->first == 3) {
      ans.insert(ans.end(), it->second, 3);
      factorial[2] -= it->second;
    } else if (it->first == 5) {
      ans.insert(ans.end(), it->second, 5);
      factorial[3] -= it->second;
      factorial[2] -= it->second * 3;
    } else if (it->first == 7) {
      ans.insert(ans.end(), it->second, 7);
      factorial[5] -= it->second;
      factorial[3] -= it->second * 2;
      factorial[2] -= it->second * 4;
    }
  }
  for (auto i : ans) cout << i;
}

### Prompt

Please formulate 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 main() {
  int t;
  cin >> t;
  long long no;
  cin >> no;
  list<int> prime_nos = {2, 3, 5, 7}, pn = prime_nos;
  map<int, int> factorial;
  while (no != 0) {
    long long n = no % 10;
    if (n == 2)
      factorial[2]++;
    else if (n == 3)
      factorial[3]++, factorial[2]++;
    else if (n == 4)
      factorial[3]++, factorial[2] += 3;
    else if (n == 5)
      factorial[5]++, factorial[3]++, factorial[2] += 3;
    else if (n == 6)
      factorial[5]++, factorial[3] += 2, factorial[2] += 4;
    else if (n == 7)
      factorial[7]++, factorial[5]++, factorial[3] += 2, factorial[2] += 4;
    else if (n == 8)
      factorial[7]++, factorial[5]++, factorial[3] += 2, factorial[2] += 7;
    else if (n == 9)
      factorial[7]++, factorial[5]++, factorial[3] += 4, factorial[2] += 7;
    no /= 10;
  }
  list<int> ans;
  for (auto it = factorial.rbegin(); it != factorial.rend(); it++) {
    if (it->first == 2) {
      ans.insert(ans.end(), it->second, 2);
    } else if (it->first == 3) {
      ans.insert(ans.end(), it->second, 3);
      factorial[2] -= it->second;
    } else if (it->first == 5) {
      ans.insert(ans.end(), it->second, 5);
      factorial[3] -= it->second;
      factorial[2] -= it->second * 3;
    } else if (it->first == 7) {
      ans.insert(ans.end(), it->second, 7);
      factorial[5] -= it->second;
      factorial[3] -= it->second * 2;
      factorial[2] -= it->second * 4;
    }
  }
  for (auto i : ans) cout << i;
}

```