#include <bits/stdc++.h>
int main(void) {
  int list[4] = {0, 0, 0, 0}, print_list[4] = {2, 3, 5, 7};
  int n, i, j;
  scanf("%d", &n);
  char c;
  for (i = 0; i < n; i++) {
    scanf(" %c", &c);
    switch (c) {
      case '9':
        list[1]++;
        list[1]++;
      case '8':
        list[0]++;
        list[0]++;
        list[0]++;
      case '7':
        list[3]++;
      case '6':
        list[1]++;
        list[0]++;
      case '5':
        list[2]++;
      case '4':
        list[0]++;
        list[0]++;
      case '3':
        list[1]++;
      case '2':
        list[0]++;
      case '0':
      case '1':
        break;
    }
  }
  for (i = 9; i >= 2; i--) {
    int flag;
    do {
      flag = 0;
      switch (i) {
        case 7:
          if (list[0] >= 4 && list[1] >= 2 && list[2] >= 1 && list[3] >= 1) {
            list[0] -= 4;
            list[1] -= 2;
            list[2] -= 1;
            list[3] -= 1;
            printf("7");
            flag = 1;
          }
          break;
        case 5:
          if (list[0] >= 3 && list[1] >= 1 && list[2] >= 1) {
            list[0] -= 3;
            list[1] -= 1;
            list[2] -= 1;
            printf("5");
            flag = 1;
          }
          break;
        case 3:
          if (list[0] >= 1 && list[1] >= 1) {
            list[0] -= 1;
            list[1] -= 1;
            printf("3");
            flag = 1;
          }
          break;
        case 2:
          if (list[0] >= 1) {
            list[0] -= 1;
            printf("2");
            flag = 1;
          }
          break;
        case '0':
        case '1':
          break;
      }
    } while (flag == 1);
  }
  return (0);
}

### 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>
int main(void) {
  int list[4] = {0, 0, 0, 0}, print_list[4] = {2, 3, 5, 7};
  int n, i, j;
  scanf("%d", &n);
  char c;
  for (i = 0; i < n; i++) {
    scanf(" %c", &c);
    switch (c) {
      case '9':
        list[1]++;
        list[1]++;
      case '8':
        list[0]++;
        list[0]++;
        list[0]++;
      case '7':
        list[3]++;
      case '6':
        list[1]++;
        list[0]++;
      case '5':
        list[2]++;
      case '4':
        list[0]++;
        list[0]++;
      case '3':
        list[1]++;
      case '2':
        list[0]++;
      case '0':
      case '1':
        break;
    }
  }
  for (i = 9; i >= 2; i--) {
    int flag;
    do {
      flag = 0;
      switch (i) {
        case 7:
          if (list[0] >= 4 && list[1] >= 2 && list[2] >= 1 && list[3] >= 1) {
            list[0] -= 4;
            list[1] -= 2;
            list[2] -= 1;
            list[3] -= 1;
            printf("7");
            flag = 1;
          }
          break;
        case 5:
          if (list[0] >= 3 && list[1] >= 1 && list[2] >= 1) {
            list[0] -= 3;
            list[1] -= 1;
            list[2] -= 1;
            printf("5");
            flag = 1;
          }
          break;
        case 3:
          if (list[0] >= 1 && list[1] >= 1) {
            list[0] -= 1;
            list[1] -= 1;
            printf("3");
            flag = 1;
          }
          break;
        case 2:
          if (list[0] >= 1) {
            list[0] -= 1;
            printf("2");
            flag = 1;
          }
          break;
        case '0':
        case '1':
          break;
      }
    } while (flag == 1);
  }
  return (0);
}

```