#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, a[10] = {0};
  string s;
  cin >> n;
  cin >> s;
  for (int i = 0; i < n; i++)
    for (int j = 2; j <= s[i] - 48; j++)
      if (j == 4)
        a[2] += 2;
      else if (j == 6) {
        a[3]++;
        a[2]++;
      } else if (j == 8)
        a[2] += 3;
      else if (j == 9)
        a[3] += 2;
      else
        a[j]++;
  while (a[2] > 0) {
    if (a[7] > 0) {
      a[2] -= 4;
      a[3] -= 2;
      a[5]--;
      a[7]--;
      cout << 7;
    } else if (a[5] > 0) {
      a[2] -= 3;
      a[3]--;
      a[5]--;
      cout << 5;
    } else if (a[3] > 0) {
      a[2]--;
      a[3]--;
      cout << 3;
    } else {
      a[2]--;
      cout << 2;
    }
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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, a[10] = {0};
  string s;
  cin >> n;
  cin >> s;
  for (int i = 0; i < n; i++)
    for (int j = 2; j <= s[i] - 48; j++)
      if (j == 4)
        a[2] += 2;
      else if (j == 6) {
        a[3]++;
        a[2]++;
      } else if (j == 8)
        a[2] += 3;
      else if (j == 9)
        a[3] += 2;
      else
        a[j]++;
  while (a[2] > 0) {
    if (a[7] > 0) {
      a[2] -= 4;
      a[3] -= 2;
      a[5]--;
      a[7]--;
      cout << 7;
    } else if (a[5] > 0) {
      a[2] -= 3;
      a[3]--;
      a[5]--;
      cout << 5;
    } else if (a[3] > 0) {
      a[2]--;
      a[3]--;
      cout << 3;
    } else {
      a[2]--;
      cout << 2;
    }
  }
  return 0;
}

```