#include <bits/stdc++.h>
using namespace std;
void setup(vector<vector<int>> &v) {
  v[0] = {2};
  v[1] = {3};
  v[2] = {2, 2, 3};
  v[3] = {5};
  v[4] = {5, 3};
  v[5] = {7};
  v[6] = {7, 2, 2, 2};
  v[7] = {7, 3, 3, 2};
}
int main() {
  int k, n, i, j;
  vector<vector<int>> v(8);
  vector<int> num;
  setup(v);
  string str;
  cin >> n >> str;
  for (i = 0; i < n; i++) {
    j = str[i] - '2';
    if (j == -2 || j == -1) continue;
    for (k = 0; k < v[j].size(); k++) {
      num.push_back(v[j][k]);
    }
  }
  sort(num.rbegin(), num.rend());
  for (i = 0; i < num.size(); i++) {
    cout << num[i];
  }
  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;
void setup(vector<vector<int>> &v) {
  v[0] = {2};
  v[1] = {3};
  v[2] = {2, 2, 3};
  v[3] = {5};
  v[4] = {5, 3};
  v[5] = {7};
  v[6] = {7, 2, 2, 2};
  v[7] = {7, 3, 3, 2};
}
int main() {
  int k, n, i, j;
  vector<vector<int>> v(8);
  vector<int> num;
  setup(v);
  string str;
  cin >> n >> str;
  for (i = 0; i < n; i++) {
    j = str[i] - '2';
    if (j == -2 || j == -1) continue;
    for (k = 0; k < v[j].size(); k++) {
      num.push_back(v[j][k]);
    }
  }
  sort(num.rbegin(), num.rend());
  for (i = 0; i < num.size(); i++) {
    cout << num[i];
  }
  return 0;
}

```