#include <bits/stdc++.h>
using namespace std;
const int INF = 10e9 + 10;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  long long a;
  cin >> a;
  vector<int> arr(n, 0);
  int l = 0;
  while (a > 0) {
    arr[l] = a % 10;
    a = a / 10;
    if (arr[l] == 0 || arr[l] == 1) {
      continue;
    }
    l++;
  }
  vector<int> ans(5 * l, 0);
  int loc = 0;
  for (int i = 0; i < l; i++) {
    switch (arr[i]) {
      case 9:
        ans[loc] = 7;
        loc++;
        ans[loc] = 3;
        loc++;
        ans[loc] = 3;
        loc++;
        ans[loc] = 2;
        loc++;
        break;
      case 8:
        ans[loc] = 7;
        loc++;
        ans[loc] = 2;
        loc++;
        ans[loc] = 2;
        loc++;
        ans[loc] = 2;
        loc++;
        break;
      case 7:
        ans[loc] = 7;
        loc++;
        break;
      case 6:
        ans[loc] = 5;
        loc++;
        ans[loc] = 3;
        loc++;
        break;
      case 5:
        ans[loc] = 5;
        loc++;
        break;
      case 4:
        ans[loc] = 3;
        loc++;
        ans[loc] = 2;
        loc++;
        ans[loc] = 2;
        loc++;
        break;
      case 3:
        ans[loc] = 3;
        loc++;
        break;
      case 2:
        ans[loc] = 2;
        loc++;
    }
  }
  sort(ans.rbegin(), ans.rend());
  for (int i = 0; i < 5 * l; i++) {
    if (ans[i] == 0) break;
    cout << ans[i];
  }
  cout << '\n';
  return 0;
}

### Prompt

Generate 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;
const int INF = 10e9 + 10;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  long long a;
  cin >> a;
  vector<int> arr(n, 0);
  int l = 0;
  while (a > 0) {
    arr[l] = a % 10;
    a = a / 10;
    if (arr[l] == 0 || arr[l] == 1) {
      continue;
    }
    l++;
  }
  vector<int> ans(5 * l, 0);
  int loc = 0;
  for (int i = 0; i < l; i++) {
    switch (arr[i]) {
      case 9:
        ans[loc] = 7;
        loc++;
        ans[loc] = 3;
        loc++;
        ans[loc] = 3;
        loc++;
        ans[loc] = 2;
        loc++;
        break;
      case 8:
        ans[loc] = 7;
        loc++;
        ans[loc] = 2;
        loc++;
        ans[loc] = 2;
        loc++;
        ans[loc] = 2;
        loc++;
        break;
      case 7:
        ans[loc] = 7;
        loc++;
        break;
      case 6:
        ans[loc] = 5;
        loc++;
        ans[loc] = 3;
        loc++;
        break;
      case 5:
        ans[loc] = 5;
        loc++;
        break;
      case 4:
        ans[loc] = 3;
        loc++;
        ans[loc] = 2;
        loc++;
        ans[loc] = 2;
        loc++;
        break;
      case 3:
        ans[loc] = 3;
        loc++;
        break;
      case 2:
        ans[loc] = 2;
        loc++;
    }
  }
  sort(ans.rbegin(), ans.rend());
  for (int i = 0; i < 5 * l; i++) {
    if (ans[i] == 0) break;
    cout << ans[i];
  }
  cout << '\n';
  return 0;
}

```