#include <bits/stdc++.h>
using namespace std;
char s[20];
int n, p[10];
int main() {
  scanf("%d", &n);
  scanf("%s", s);
  int l = strlen(s);
  for (int i = 0; i < l; i++) {
    int j = s[i] - 48;
    for (int np = j; np; np--) {
      int r = np;
      while (np % 2 == 0) {
        p[2]++;
        np /= 2;
      }
      while (np % 3 == 0) {
        p[3]++;
        np /= 3;
      }
      while (np % 5 == 0) {
        p[5]++;
        np /= 5;
      }
      while (np % 7 == 0) {
        p[7]++;
        np /= 7;
      }
      np = r;
    }
  }
  while (p[7]) {
    p[7]--;
    p[5]--;
    p[3] -= 2;
    p[2] -= 4;
    printf("7");
  }
  while (p[5]) {
    p[5]--;
    p[3]--;
    p[2] -= 3;
    printf("5");
  }
  while (p[3]) {
    p[3]--;
    p[2]--;
    printf("3");
  }
  while (p[2]) {
    p[2]--;
    printf("2");
  }
  printf("\n");
  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>
using namespace std;
char s[20];
int n, p[10];
int main() {
  scanf("%d", &n);
  scanf("%s", s);
  int l = strlen(s);
  for (int i = 0; i < l; i++) {
    int j = s[i] - 48;
    for (int np = j; np; np--) {
      int r = np;
      while (np % 2 == 0) {
        p[2]++;
        np /= 2;
      }
      while (np % 3 == 0) {
        p[3]++;
        np /= 3;
      }
      while (np % 5 == 0) {
        p[5]++;
        np /= 5;
      }
      while (np % 7 == 0) {
        p[7]++;
        np /= 7;
      }
      np = r;
    }
  }
  while (p[7]) {
    p[7]--;
    p[5]--;
    p[3] -= 2;
    p[2] -= 4;
    printf("7");
  }
  while (p[5]) {
    p[5]--;
    p[3]--;
    p[2] -= 3;
    printf("5");
  }
  while (p[3]) {
    p[3]--;
    p[2]--;
    printf("3");
  }
  while (p[2]) {
    p[2]--;
    printf("2");
  }
  printf("\n");
  return 0;
}

```