#include <bits/stdc++.h>
using namespace std;
int factorPow(int& n, int d) {
  int cnt = 0;
  while (n > 0 && n % d == 0) {
    n /= d;
    cnt++;
  }
  return cnt;
}
void solve() {
  int n;
  cin >> n;
  string a;
  cin >> a;
  vector<int> digits(n);
  for (int i = 0; i < n; i++) {
    digits[i] = a[i] - '0';
  }
  vector<int> fact(10);
  fact[0] = 1;
  for (int i = 1; i <= 9; i++) {
    fact[i] = i * fact[i - 1];
  }
  vector<int> primes{2, 3, 5, 7};
  string ans = "";
  for (int i = 0; i < n; i++) {
    int f = fact[digits[i]];
    for (int j = 3; j >= 0; j--) {
      ans.append(factorPow(f, fact[primes[j]]), primes[j] + '0');
    }
  }
  sort(ans.begin(), ans.end());
  reverse(ans.begin(), ans.end());
  cout << ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return EXIT_SUCCESS;
}

### 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;
int factorPow(int& n, int d) {
  int cnt = 0;
  while (n > 0 && n % d == 0) {
    n /= d;
    cnt++;
  }
  return cnt;
}
void solve() {
  int n;
  cin >> n;
  string a;
  cin >> a;
  vector<int> digits(n);
  for (int i = 0; i < n; i++) {
    digits[i] = a[i] - '0';
  }
  vector<int> fact(10);
  fact[0] = 1;
  for (int i = 1; i <= 9; i++) {
    fact[i] = i * fact[i - 1];
  }
  vector<int> primes{2, 3, 5, 7};
  string ans = "";
  for (int i = 0; i < n; i++) {
    int f = fact[digits[i]];
    for (int j = 3; j >= 0; j--) {
      ans.append(factorPow(f, fact[primes[j]]), primes[j] + '0');
    }
  }
  sort(ans.begin(), ans.end());
  reverse(ans.begin(), ans.end());
  cout << ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return EXIT_SUCCESS;
}

```