#include <bits/stdc++.h>
using namespace std;
bool cmp(char a, char b) { return a > b; }
int main() {
  int N, j = 0;
  char A[16], B[100];
  scanf("%d\n%s", &N, A);
  for (int i = 0; i < N; i++) {
    if (A[i] == '2' || A[i] == '3' || A[i] == '5' || A[i] == '7') {
      B[j++] = A[i];
    } else if (A[i] == '4') {
      B[j++] = '2';
      B[j++] = '2';
      B[j++] = '3';
    } else if (A[i] == '6') {
      B[j++] = '3';
      B[j++] = '5';
    } else if (A[i] == '8') {
      B[j++] = '2';
      B[j++] = '2';
      B[j++] = '2';
      B[j++] = '7';
    } else if (A[i] == '9') {
      B[j++] = '2';
      B[j++] = '3';
      B[j++] = '3';
      B[j++] = '7';
    }
  }
  sort(B, B + j, cmp);
  for (int i = 0; i < j; i++) {
    printf("%c", B[i]);
  }
  printf("\n");
}

### 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;
bool cmp(char a, char b) { return a > b; }
int main() {
  int N, j = 0;
  char A[16], B[100];
  scanf("%d\n%s", &N, A);
  for (int i = 0; i < N; i++) {
    if (A[i] == '2' || A[i] == '3' || A[i] == '5' || A[i] == '7') {
      B[j++] = A[i];
    } else if (A[i] == '4') {
      B[j++] = '2';
      B[j++] = '2';
      B[j++] = '3';
    } else if (A[i] == '6') {
      B[j++] = '3';
      B[j++] = '5';
    } else if (A[i] == '8') {
      B[j++] = '2';
      B[j++] = '2';
      B[j++] = '2';
      B[j++] = '7';
    } else if (A[i] == '9') {
      B[j++] = '2';
      B[j++] = '3';
      B[j++] = '3';
      B[j++] = '7';
    }
  }
  sort(B, B + j, cmp);
  for (int i = 0; i < j; i++) {
    printf("%c", B[i]);
  }
  printf("\n");
}

```