#include <bits/stdc++.h>
using namespace std;
void display(std::vector<int> v) {
  for (int i = 3; i >= 0; i--) {
    for (int j = 0; j < v[i]; j++) {
      if (i == 3)
        cout << 7;
      else if (i == 2)
        cout << 5;
      else if (i == 1)
        cout << 3;
      else if (i == 0)
        cout << 2;
    }
  }
}
int main() {
  int n;
  cin >> n;
  std::vector<int> v = {0, 0, 0, 0};
  while (n--) {
    char c;
    cin >> c;
    if (c == '2')
      v[0]++;
    else if (c == '3')
      v[1]++;
    else if (c == '4') {
      v[0] += 2;
      v[1]++;
    } else if (c == '5')
      v[2]++;
    else if (c == '6') {
      v[1] += 1;
      v[2]++;
    } else if (c == '7')
      v[3]++;
    else if (c == '8') {
      v[3]++;
      v[0] += 3;
    } else if (c == '9') {
      v[1] += 2;
      v[0]++;
      v[3]++;
    }
  }
  display(v);
  return 0;
}

### 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;
void display(std::vector<int> v) {
  for (int i = 3; i >= 0; i--) {
    for (int j = 0; j < v[i]; j++) {
      if (i == 3)
        cout << 7;
      else if (i == 2)
        cout << 5;
      else if (i == 1)
        cout << 3;
      else if (i == 0)
        cout << 2;
    }
  }
}
int main() {
  int n;
  cin >> n;
  std::vector<int> v = {0, 0, 0, 0};
  while (n--) {
    char c;
    cin >> c;
    if (c == '2')
      v[0]++;
    else if (c == '3')
      v[1]++;
    else if (c == '4') {
      v[0] += 2;
      v[1]++;
    } else if (c == '5')
      v[2]++;
    else if (c == '6') {
      v[1] += 1;
      v[2]++;
    } else if (c == '7')
      v[3]++;
    else if (c == '8') {
      v[3]++;
      v[0] += 3;
    } else if (c == '9') {
      v[1] += 2;
      v[0]++;
      v[3]++;
    }
  }
  display(v);
  return 0;
}

```