#include <bits/stdc++.h>
using namespace std;
const long double INF = 1e18 + 7;
long long N = 1e6 * 5;
long double EPS = 1 / 1e18;
long double PI = 3.14159265358979323846;
long long first[8] = {2, 3, 322, 5, 53, 7, 7222, 7332}, ans[70];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, cnt = 0;
  cin >> n;
  string second;
  cin >> second;
  for (long long i = 0; i < n; i++) {
    if (second[i] - '0' >= 2) {
      long long num = first[second[i] - '2'];
      for (; num > 0;) {
        ans[cnt++] = num % 10;
        num /= 10;
      }
    }
  }
  sort(ans, ans + cnt);
  for (long long i = cnt - 1; i >= 0; i--) cout << ans[i];
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
const long double INF = 1e18 + 7;
long long N = 1e6 * 5;
long double EPS = 1 / 1e18;
long double PI = 3.14159265358979323846;
long long first[8] = {2, 3, 322, 5, 53, 7, 7222, 7332}, ans[70];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, cnt = 0;
  cin >> n;
  string second;
  cin >> second;
  for (long long i = 0; i < n; i++) {
    if (second[i] - '0' >= 2) {
      long long num = first[second[i] - '2'];
      for (; num > 0;) {
        ans[cnt++] = num % 10;
        num /= 10;
      }
    }
  }
  sort(ans, ans + cnt);
  for (long long i = cnt - 1; i >= 0; i--) cout << ans[i];
  return 0;
}

```