#include <bits/stdc++.h>
int main() {
  short n;
  std::cin >> n;
  int a[10] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
  std::string str;
  std::cin >> str;
  for (int i = 0; i < n; ++i) ++a[(int)str[i] - 48];
  int ans[8] = {0, 0, 0, 0, 0, 0, 0, 0};
  while (a[9]-- > 0) {
    ++ans[7];
    ans[3] += 2;
    ++ans[2];
  }
  while (a[8]-- > 0) {
    ++ans[7];
    ans[2] += 3;
  }
  ans[7] += a[7];
  while (a[6]-- > 0) {
    ++ans[5];
    ++ans[3];
  }
  ans[5] += a[5];
  while (a[4]-- > 0) {
    ++ans[3];
    ans[2] += 2;
  }
  ans[3] += a[3];
  ans[2] += a[2];
  for (int i = 7; i > 1; --i)
    while (ans[i] > 0) {
      std::cout << i;
      --ans[i];
    }
  return 0;
}

### 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>
int main() {
  short n;
  std::cin >> n;
  int a[10] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
  std::string str;
  std::cin >> str;
  for (int i = 0; i < n; ++i) ++a[(int)str[i] - 48];
  int ans[8] = {0, 0, 0, 0, 0, 0, 0, 0};
  while (a[9]-- > 0) {
    ++ans[7];
    ans[3] += 2;
    ++ans[2];
  }
  while (a[8]-- > 0) {
    ++ans[7];
    ans[2] += 3;
  }
  ans[7] += a[7];
  while (a[6]-- > 0) {
    ++ans[5];
    ++ans[3];
  }
  ans[5] += a[5];
  while (a[4]-- > 0) {
    ++ans[3];
    ans[2] += 2;
  }
  ans[3] += a[3];
  ans[2] += a[2];
  for (int i = 7; i > 1; --i)
    while (ans[i] > 0) {
      std::cout << i;
      --ans[i];
    }
  return 0;
}

```