#include <bits/stdc++.h>
using namespace std;
int main() {
  string s;
  int n;
  cin >> n;
  cin >> s;
  vector<int> count(10, 0);
  for (int i = 0; i < s.size(); ++i) {
    int k = s[i] - '0';
    for (int j = 0; j < k + 1; ++j) {
      count[j]++;
    }
  }
  for (int i = 9; i >= 2; --i) {
    switch (i) {
      case 9:
        count[3] += count[9] * 2;
        count[9] = 0;
        break;
      case 8:
        count[2] += count[8] * 3;
        count[8] = 0;
        break;
      case 6:
        count[2] += count[6];
        count[3] += count[6];
        count[6] = 0;
        break;
      case 4:
        count[2] += count[4] * 2;
        count[4] = 0;
        break;
      default:
        break;
    }
    while (count[i] > 0) {
      cout << i;
      for (int l = i; l >= 2; --l) {
        count[l]--;
      }
    }
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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() {
  string s;
  int n;
  cin >> n;
  cin >> s;
  vector<int> count(10, 0);
  for (int i = 0; i < s.size(); ++i) {
    int k = s[i] - '0';
    for (int j = 0; j < k + 1; ++j) {
      count[j]++;
    }
  }
  for (int i = 9; i >= 2; --i) {
    switch (i) {
      case 9:
        count[3] += count[9] * 2;
        count[9] = 0;
        break;
      case 8:
        count[2] += count[8] * 3;
        count[8] = 0;
        break;
      case 6:
        count[2] += count[6];
        count[3] += count[6];
        count[6] = 0;
        break;
      case 4:
        count[2] += count[4] * 2;
        count[4] = 0;
        break;
      default:
        break;
    }
    while (count[i] > 0) {
      cout << i;
      for (int l = i; l >= 2; --l) {
        count[l]--;
      }
    }
  }
  return 0;
}

```