#include <bits/stdc++.h>
using namespace std;
vector<int> v;
int main() {
  long long int i, j, k, cs, t;
  cin >> t;
  cin >> k;
  int A[10][10] = {
      {-1},    {-1},       {2, -1}, {3, -1},          {2, 2, 3, -1},
      {5, -1}, {5, 3, -1}, {7, -1}, {7, 2, 2, 2, -1}, {7, 3, 3, 2, -1}};
  while (k) {
    i = k % 10;
    for (j = 0;; j++) {
      if (A[i][j] == -1) break;
      v.push_back(A[i][j]);
    }
    k /= 10;
  }
  sort(v.begin(), v.end());
  reverse(v.begin(), v.end());
  for (i = 0; i < v.size(); i++) cout << v[i];
  cout << endl;
  return 0;
}

### Prompt

Please formulate 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;
vector<int> v;
int main() {
  long long int i, j, k, cs, t;
  cin >> t;
  cin >> k;
  int A[10][10] = {
      {-1},    {-1},       {2, -1}, {3, -1},          {2, 2, 3, -1},
      {5, -1}, {5, 3, -1}, {7, -1}, {7, 2, 2, 2, -1}, {7, 3, 3, 2, -1}};
  while (k) {
    i = k % 10;
    for (j = 0;; j++) {
      if (A[i][j] == -1) break;
      v.push_back(A[i][j]);
    }
    k /= 10;
  }
  sort(v.begin(), v.end());
  reverse(v.begin(), v.end());
  for (i = 0; i < v.size(); i++) cout << v[i];
  cout << endl;
  return 0;
}

```