#include <bits/stdc++.h>
long long int number, digitnumber, digit[20];
int numb[11];
int main() {
  scanf("%lld", &digitnumber);
  scanf("%lld", &number);
  for (int i = 1; i <= digitnumber; i++) {
    numb[number % 10]++;
    number /= 10;
  }
  while (numb[4] != 0) {
    numb[4]--;
    numb[2] += 2;
    numb[3]++;
  }
  while (numb[6] != 0) {
    numb[6]--;
    numb[3]++;
    numb[5]++;
  }
  while (numb[8] != 0) {
    numb[8]--;
    numb[2] += 3;
    numb[7]++;
  }
  while (numb[9] != 0) {
    numb[9]--;
    numb[2]++;
    numb[3] += 2;
    numb[7]++;
  }
  for (int i = 7; i >= 2; i--)
    for (int j = 1; j <= numb[i]; j++) printf("%d", i);
}

### Prompt

Please provide a CPP coded solution 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>
long long int number, digitnumber, digit[20];
int numb[11];
int main() {
  scanf("%lld", &digitnumber);
  scanf("%lld", &number);
  for (int i = 1; i <= digitnumber; i++) {
    numb[number % 10]++;
    number /= 10;
  }
  while (numb[4] != 0) {
    numb[4]--;
    numb[2] += 2;
    numb[3]++;
  }
  while (numb[6] != 0) {
    numb[6]--;
    numb[3]++;
    numb[5]++;
  }
  while (numb[8] != 0) {
    numb[8]--;
    numb[2] += 3;
    numb[7]++;
  }
  while (numb[9] != 0) {
    numb[9]--;
    numb[2]++;
    numb[3] += 2;
    numb[7]++;
  }
  for (int i = 7; i >= 2; i--)
    for (int j = 1; j <= numb[i]; j++) printf("%d", i);
}

```