#include <bits/stdc++.h>
using namespace std;
int ans[10];
int main() {
  memset(ans, 0, sizeof(ans));
  int n;
  char c;
  scanf("%d", &n);
  while (n--) {
    scanf(" %c", &c);
    if (c == '9') {
      ans[7]++;
      ans[3] += 2;
      ans[2]++;
    }
    if (c == '8') {
      ans[7]++;
      ans[2] += 3;
    }
    if (c == '7') ans[7]++;
    if (c == '6') {
      ans[5]++;
      ans[3]++;
    }
    if (c == '5') ans[5]++;
    if (c == '4') {
      ans[3]++;
      ans[2] += 2;
    }
    if (c == '3') ans[3]++;
    if (c == '2') ans[2]++;
  }
  for (int i = 9; i > 1; i--)
    for (int j = 0; j < ans[i]; j++) printf("%d", i);
  printf("\n");
  return 0;
}

### Prompt

In cpp, your task is to solve 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>
using namespace std;
int ans[10];
int main() {
  memset(ans, 0, sizeof(ans));
  int n;
  char c;
  scanf("%d", &n);
  while (n--) {
    scanf(" %c", &c);
    if (c == '9') {
      ans[7]++;
      ans[3] += 2;
      ans[2]++;
    }
    if (c == '8') {
      ans[7]++;
      ans[2] += 3;
    }
    if (c == '7') ans[7]++;
    if (c == '6') {
      ans[5]++;
      ans[3]++;
    }
    if (c == '5') ans[5]++;
    if (c == '4') {
      ans[3]++;
      ans[2] += 2;
    }
    if (c == '3') ans[3]++;
    if (c == '2') ans[2]++;
  }
  for (int i = 9; i > 1; i--)
    for (int j = 0; j < ans[i]; j++) printf("%d", i);
  printf("\n");
  return 0;
}

```