#include <bits/stdc++.h>
using namespace std;
void solve(long long testcase) {
  long long n;
  cin >> n;
  string s;
  cin >> s;
  vector<long long> v[10] = {{},  {},     {2}, {3},          {3, 2, 2},
                             {5}, {5, 3}, {7}, {2, 2, 2, 7}, {7, 3, 3, 2}},
                    ans;
  for (char c : s) {
    long long d = c - '0';
    for (long long x : v[d]) ans.push_back(x);
  }
  sort(ans.begin(), ans.end(), greater<long long>());
  for (long long x : ans) cout << x;
  cout << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T = 1;
  for (long long t = 1; t <= T; t++) solve(t);
  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 solve(long long testcase) {
  long long n;
  cin >> n;
  string s;
  cin >> s;
  vector<long long> v[10] = {{},  {},     {2}, {3},          {3, 2, 2},
                             {5}, {5, 3}, {7}, {2, 2, 2, 7}, {7, 3, 3, 2}},
                    ans;
  for (char c : s) {
    long long d = c - '0';
    for (long long x : v[d]) ans.push_back(x);
  }
  sort(ans.begin(), ans.end(), greater<long long>());
  for (long long x : ans) cout << x;
  cout << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T = 1;
  for (long long t = 1; t <= T; t++) solve(t);
  return 0;
}

```