#include <bits/stdc++.h>
using namespace std;
int dig[10];
void addDig(int x) {
  switch (x) {
    case 4:
      dig[3]++;
      dig[2] += 2;
      break;
    case 6:
      dig[5]++;
      dig[3]++;
      break;
    case 8:
      dig[7]++;
      dig[2] += 3;
      break;
    case 9:
      dig[7]++;
      dig[3] += 2;
      dig[2]++;
      break;
    default:
      dig[x]++;
  }
}
int main() {
  long long n;
  cin >> n >> n;
  while (n > 0) {
    addDig(n % 10);
    n /= 10;
  }
  for (int i = 9; i > 1; i--)
    for (int j = 0; j < dig[i]; j++) cout << i;
  cout << '\n';
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 dig[10];
void addDig(int x) {
  switch (x) {
    case 4:
      dig[3]++;
      dig[2] += 2;
      break;
    case 6:
      dig[5]++;
      dig[3]++;
      break;
    case 8:
      dig[7]++;
      dig[2] += 3;
      break;
    case 9:
      dig[7]++;
      dig[3] += 2;
      dig[2]++;
      break;
    default:
      dig[x]++;
  }
}
int main() {
  long long n;
  cin >> n >> n;
  while (n > 0) {
    addDig(n % 10);
    n /= 10;
  }
  for (int i = 9; i > 1; i--)
    for (int j = 0; j < dig[i]; j++) cout << i;
  cout << '\n';
  return 0;
}

```