#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  string s;
  cin >> n;
  cin >> s;
  map<int, int> xxx;
  n = s.length();
  int tmp2;
  for (int i = 0; i < n; i++) {
    tmp2 = s[i] - 48;
    if (tmp2 == 4) {
      xxx[2] += 2;
      xxx[3]++;
    } else if (tmp2 == 6) {
      xxx[5]++;
      xxx[3]++;
    } else if (tmp2 == 8) {
      xxx[2] += 3;
      xxx[7]++;
    } else if (tmp2 == 9) {
      xxx[7]++;
      xxx[3] += 2;
      xxx[2]++;
    } else {
      xxx[tmp2]++;
    }
  }
  int tmp;
  for (int i = 9; i >= 2; i--) {
    tmp = xxx[i];
    for (int j = 1; j <= tmp; j++) cout << i;
  }
  cout << endl;
}

### 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 main() {
  int n;
  string s;
  cin >> n;
  cin >> s;
  map<int, int> xxx;
  n = s.length();
  int tmp2;
  for (int i = 0; i < n; i++) {
    tmp2 = s[i] - 48;
    if (tmp2 == 4) {
      xxx[2] += 2;
      xxx[3]++;
    } else if (tmp2 == 6) {
      xxx[5]++;
      xxx[3]++;
    } else if (tmp2 == 8) {
      xxx[2] += 3;
      xxx[7]++;
    } else if (tmp2 == 9) {
      xxx[7]++;
      xxx[3] += 2;
      xxx[2]++;
    } else {
      xxx[tmp2]++;
    }
  }
  int tmp;
  for (int i = 9; i >= 2; i--) {
    tmp = xxx[i];
    for (int j = 1; j <= tmp; j++) cout << i;
  }
  cout << endl;
}

```