#include <bits/stdc++.h>
using namespace std;
int main() {
  int e[10] = {0}, n;
  char a;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a;
    if (a == 50) e[2]++;
    if (a == 51) e[3]++;
    if (a == 52) {
      e[2] += 2;
      e[3]++;
    }
    if (a == 53) e[5]++;
    if (a == 54) {
      e[3]++;
      e[5]++;
    }
    if (a == 55) e[7]++;
    if (a == 56) {
      e[2] += 3;
      e[7]++;
    }
    if (a == 57) {
      e[3] += 2;
      e[2]++;
      e[7]++;
    }
  }
  for (int c = 9; c >= 2; c--)
    for (int d = 0; d < e[c]; d++) cout << c;
  cout << endl;
}

### 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 e[10] = {0}, n;
  char a;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a;
    if (a == 50) e[2]++;
    if (a == 51) e[3]++;
    if (a == 52) {
      e[2] += 2;
      e[3]++;
    }
    if (a == 53) e[5]++;
    if (a == 54) {
      e[3]++;
      e[5]++;
    }
    if (a == 55) e[7]++;
    if (a == 56) {
      e[2] += 3;
      e[7]++;
    }
    if (a == 57) {
      e[3] += 2;
      e[2]++;
      e[7]++;
    }
  }
  for (int c = 9; c >= 2; c--)
    for (int d = 0; d < e[c]; d++) cout << c;
  cout << endl;
}

```