#include <bits/stdc++.h>
using namespace std;
int main() {
  vector<int> ans;
  vector<int> fac;
  fac.push_back(1);
  fac.push_back(1);
  for (int i = 2; i < 10; i++) fac.push_back(fac[i - 1] * i);
  int n;
  cin >> n;
  string s;
  cin >> s;
  vector<int> v;
  for (int i = 0; i < n; i++) {
    v.push_back((s[i] - '0'));
  }
  for (int i = 0; i < n; i++) {
    int x = fac[v[i]];
    vector<int> div;
    if (x % fac[7] == 0) div.push_back(7);
    if (x % fac[5] == 0) div.push_back(5);
    if (x % fac[3] == 0) div.push_back(3);
    if (x % fac[2] == 0) div.push_back(2);
    int y = 0;
    while (y != div.size()) {
      if (x == 1) break;
      if (x % fac[div[y]] == 0) {
        ans.push_back(div[y]);
        x /= fac[div[y]];
        continue;
      }
      y++;
    }
  }
  sort(ans.begin(), ans.end());
  reverse(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) cout << ans[i];
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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() {
  vector<int> ans;
  vector<int> fac;
  fac.push_back(1);
  fac.push_back(1);
  for (int i = 2; i < 10; i++) fac.push_back(fac[i - 1] * i);
  int n;
  cin >> n;
  string s;
  cin >> s;
  vector<int> v;
  for (int i = 0; i < n; i++) {
    v.push_back((s[i] - '0'));
  }
  for (int i = 0; i < n; i++) {
    int x = fac[v[i]];
    vector<int> div;
    if (x % fac[7] == 0) div.push_back(7);
    if (x % fac[5] == 0) div.push_back(5);
    if (x % fac[3] == 0) div.push_back(3);
    if (x % fac[2] == 0) div.push_back(2);
    int y = 0;
    while (y != div.size()) {
      if (x == 1) break;
      if (x % fac[div[y]] == 0) {
        ans.push_back(div[y]);
        x /= fac[div[y]];
        continue;
      }
      y++;
    }
  }
  sort(ans.begin(), ans.end());
  reverse(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) cout << ans[i];
  return 0;
}

```