#include <bits/stdc++.h>
using namespace std;
long long freq[12], minus_[12];
void precal(string s) {
  for (int i = 0; i < s.size(); i++) {
    for (int k, n = s[i] - '0'; n >= 2; n--) {
      k = n;
      for (int j = 2; j <= k; j++) {
        while (k % j == 0) {
          freq[j]++;
          k /= j;
        }
      }
    }
  }
}
void solve() {
  for (int i = 7; i >= 2; i--) {
    if (freq[i])
      while (freq[i]--) {
        cout << i;
        for (int k, n = i - 1; n >= 2; n--) {
          k = n;
          for (int j = 2; j <= k; j++) {
            while (k % j == 0) {
              freq[j]--;
              k /= j;
            }
          }
        }
      }
  }
}
int main() {
  long long n, res = 1;
  string s;
  cin >> n >> s;
  precal(s);
  solve();
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
long long freq[12], minus_[12];
void precal(string s) {
  for (int i = 0; i < s.size(); i++) {
    for (int k, n = s[i] - '0'; n >= 2; n--) {
      k = n;
      for (int j = 2; j <= k; j++) {
        while (k % j == 0) {
          freq[j]++;
          k /= j;
        }
      }
    }
  }
}
void solve() {
  for (int i = 7; i >= 2; i--) {
    if (freq[i])
      while (freq[i]--) {
        cout << i;
        for (int k, n = i - 1; n >= 2; n--) {
          k = n;
          for (int j = 2; j <= k; j++) {
            while (k % j == 0) {
              freq[j]--;
              k /= j;
            }
          }
        }
      }
  }
}
int main() {
  long long n, res = 1;
  string s;
  cin >> n >> s;
  precal(s);
  solve();
  return 0;
}

```