#include <bits/stdc++.h>
using namespace std;
char a[10][10] = {"", "", "2", "3", "322", "5", "53", "7", "7222", "7332"};
int main() {
  int n;
  char b[110], c[1000];
  while (cin >> n) {
    int cnt = 0;
    scanf("%s", b);
    for (int i = 0; i < n; i++) {
      int str = strlen(a[b[i] - '0']);
      for (int j = 0; j < str; j++) c[cnt++] = a[b[i] - '0'][j];
    }
    sort(c, c + cnt);
    for (int i = cnt - 1; i >= 0; i--) printf("%c", c[i]);
    printf("\n");
  }
  return 0;
}

### Prompt

Your task is to create 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;
char a[10][10] = {"", "", "2", "3", "322", "5", "53", "7", "7222", "7332"};
int main() {
  int n;
  char b[110], c[1000];
  while (cin >> n) {
    int cnt = 0;
    scanf("%s", b);
    for (int i = 0; i < n; i++) {
      int str = strlen(a[b[i] - '0']);
      for (int j = 0; j < str; j++) c[cnt++] = a[b[i] - '0'][j];
    }
    sort(c, c + cnt);
    for (int i = cnt - 1; i >= 0; i--) printf("%c", c[i]);
    printf("\n");
  }
  return 0;
}

```