#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  vector<int> nums(10, 0);
  char temp;
  for (int i = 0; i < n; i++) {
    cin >> temp;
    int s = temp - '0';
    for (int j = 1; j <= s; j++) {
      nums[j] += 1;
    }
  }
  for (int i = 0; i < nums[7]; i++) {
    cout << 7;
    for (int j = 0; j < 7; j++) {
      nums[j]--;
    }
  }
  for (int i = 0; i < nums[5]; i++) {
    cout << 5;
    for (int j = 0; j < 5; j++) {
      nums[j]--;
    }
  }
  nums[2] += nums[4] * 2;
  nums[2] += nums[8] * 3;
  nums[3] += nums[9] * 2;
  nums[3] += nums[6];
  nums[2] += nums[6];
  nums[4] = 0;
  nums[6] = 0;
  nums[8] = 0;
  nums[9] = 0;
  for (int i = 0; i < nums[3]; i++) {
    cout << 3;
    for (int j = 0; j < 3; j++) {
      nums[j]--;
    }
  }
  for (int i = 0; i < nums[2]; i++) {
    cout << 2;
  }
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
int main() {
  int n;
  cin >> n;
  vector<int> nums(10, 0);
  char temp;
  for (int i = 0; i < n; i++) {
    cin >> temp;
    int s = temp - '0';
    for (int j = 1; j <= s; j++) {
      nums[j] += 1;
    }
  }
  for (int i = 0; i < nums[7]; i++) {
    cout << 7;
    for (int j = 0; j < 7; j++) {
      nums[j]--;
    }
  }
  for (int i = 0; i < nums[5]; i++) {
    cout << 5;
    for (int j = 0; j < 5; j++) {
      nums[j]--;
    }
  }
  nums[2] += nums[4] * 2;
  nums[2] += nums[8] * 3;
  nums[3] += nums[9] * 2;
  nums[3] += nums[6];
  nums[2] += nums[6];
  nums[4] = 0;
  nums[6] = 0;
  nums[8] = 0;
  nums[9] = 0;
  for (int i = 0; i < nums[3]; i++) {
    cout << 3;
    for (int j = 0; j < 3; j++) {
      nums[j]--;
    }
  }
  for (int i = 0; i < nums[2]; i++) {
    cout << 2;
  }
  return 0;
}

```