#include <bits/stdc++.h>
using namespace std;
char s[20];
int a[500000];
int main() {
  int n;
  while (cin >> n >> s) {
    int an = 0;
    for (int i = 0; i < n; i++) {
      switch (s[i]) {
        case '2':
        case '3':
        case '5':
        case '7':
          a[an++] = s[i] - '0';
          break;
        case '4':
          a[an++] = 2;
          a[an++] = 2;
          a[an++] = 3;
          break;
        case '6':
          a[an++] = 5;
          a[an++] = 3;
          break;
        case '8':
          a[an++] = 7;
          a[an++] = 2;
          a[an++] = 2;
          a[an++] = 2;
          break;
        case '9':
          a[an++] = 7;
          a[an++] = 2;
          a[an++] = 3;
          a[an++] = 3;
          break;
      }
    }
    sort(a, a + an);
    for (int i = an - 1; i >= 0; i--) cout << a[i];
    cout << endl;
  }
}

### 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;
char s[20];
int a[500000];
int main() {
  int n;
  while (cin >> n >> s) {
    int an = 0;
    for (int i = 0; i < n; i++) {
      switch (s[i]) {
        case '2':
        case '3':
        case '5':
        case '7':
          a[an++] = s[i] - '0';
          break;
        case '4':
          a[an++] = 2;
          a[an++] = 2;
          a[an++] = 3;
          break;
        case '6':
          a[an++] = 5;
          a[an++] = 3;
          break;
        case '8':
          a[an++] = 7;
          a[an++] = 2;
          a[an++] = 2;
          a[an++] = 2;
          break;
        case '9':
          a[an++] = 7;
          a[an++] = 2;
          a[an++] = 3;
          a[an++] = 3;
          break;
      }
    }
    sort(a, a + an);
    for (int i = an - 1; i >= 0; i--) cout << a[i];
    cout << endl;
  }
}

```