#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, sz = 0;
  scanf("%d\n", &n);
  static int a[100];
  while (n--) {
    char c = getchar() - 48;
    if (c <= 1)
      continue;
    else if (c == 4)
      a[sz++] = 2, a[sz++] = 3, a[sz++] = 2;
    else if (c == 6)
      a[sz++] = 5, a[sz++] = 3;
    else if (c == 8)
      a[sz++] = 7, a[sz++] = 2, a[sz++] = 2, a[sz++] = 2;
    else if (c == 9)
      a[sz++] = 7, a[sz++] = 3, a[sz++] = 3, a[sz++] = 2;
    else
      a[sz++] = c;
  }
  sort(a, a + sz, greater<int>());
  for (int i = 0; i < sz; ++i) putchar(a[i] + 48);
}

### Prompt

Please provide a Cpp coded solution 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;
int main() {
  int n, sz = 0;
  scanf("%d\n", &n);
  static int a[100];
  while (n--) {
    char c = getchar() - 48;
    if (c <= 1)
      continue;
    else if (c == 4)
      a[sz++] = 2, a[sz++] = 3, a[sz++] = 2;
    else if (c == 6)
      a[sz++] = 5, a[sz++] = 3;
    else if (c == 8)
      a[sz++] = 7, a[sz++] = 2, a[sz++] = 2, a[sz++] = 2;
    else if (c == 9)
      a[sz++] = 7, a[sz++] = 3, a[sz++] = 3, a[sz++] = 2;
    else
      a[sz++] = c;
  }
  sort(a, a + sz, greater<int>());
  for (int i = 0; i < sz; ++i) putchar(a[i] + 48);
}

```