#include <bits/stdc++.h>
char str[20];
int a[10];
int main() {
  int n;
  while (~scanf("%d", &n)) {
    memset(a, 0, sizeof(a));
    scanf("%s", str);
    int len = strlen(str);
    for (int i = 0; i < len; i++) {
      if (str[i] == '2') {
        a[2]++;
      }
      if (str[i] == '3') {
        a[3]++;
      }
      if (str[i] == '4') {
        a[2] += 2;
        a[3]++;
      }
      if (str[i] == '5') {
        a[5]++;
      }
      if (str[i] == '6') {
        a[5]++;
        a[3]++;
      }
      if (str[i] == '7') {
        a[7]++;
      }
      if (str[i] == '8') {
        a[2] += 3;
        a[7]++;
      }
      if (str[i] == '9') {
        a[3] += 2;
        a[2]++;
        a[7]++;
      }
    }
    for (int i = 7; i >= 2; i--) {
      while (a[i]) {
        printf("%d", i);
        a[i]--;
      }
    }
    printf("\n");
  }
  return 0;
}

### Prompt

Please formulate 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>
char str[20];
int a[10];
int main() {
  int n;
  while (~scanf("%d", &n)) {
    memset(a, 0, sizeof(a));
    scanf("%s", str);
    int len = strlen(str);
    for (int i = 0; i < len; i++) {
      if (str[i] == '2') {
        a[2]++;
      }
      if (str[i] == '3') {
        a[3]++;
      }
      if (str[i] == '4') {
        a[2] += 2;
        a[3]++;
      }
      if (str[i] == '5') {
        a[5]++;
      }
      if (str[i] == '6') {
        a[5]++;
        a[3]++;
      }
      if (str[i] == '7') {
        a[7]++;
      }
      if (str[i] == '8') {
        a[2] += 3;
        a[7]++;
      }
      if (str[i] == '9') {
        a[3] += 2;
        a[2]++;
        a[7]++;
      }
    }
    for (int i = 7; i >= 2; i--) {
      while (a[i]) {
        printf("%d", i);
        a[i]--;
      }
    }
    printf("\n");
  }
  return 0;
}

```