#include <bits/stdc++.h>
using namespace std;
int factorial(int n) { return (n == 1 || n == 0) ? 1 : n * factorial(n - 1); }
void solve() {
  int digits;
  cin >> digits;
  string num;
  cin >> num;
  long long product = 1;
  string res = "";
  vector<int> counts(10, 0);
  for (int i = 0; i < digits; i++) {
    product *= factorial(num[i] - 48);
    for (int j = 1; j <= num[i] - 48; j++) {
      counts[j]++;
    }
  }
  for (int i = 1; i <= 9; i++) {
    if (i == 4) {
      while (counts[i] > 0) {
        counts[2] += 2;
        counts[i]--;
      }
    } else if (i == 6) {
      while (counts[i] > 0) {
        counts[2]++;
        counts[3]++;
        counts[i]--;
      }
    } else if (i == 8) {
      while (counts[i] > 0) {
        counts[2] += 3;
        counts[i]--;
      }
    } else if (i == 9) {
      while (counts[i] > 0) {
        counts[3] += 2;
        counts[i]--;
      }
    }
  }
  int pos = 9;
  while (pos != 1) {
    while (counts[pos] > 0) {
      res += char(pos + 48);
      if (pos == 7) {
        counts[3]--;
        counts[2]--;
        counts[2]--;
        counts[2]--;
      } else if (pos == 5) {
        counts[2]--;
        counts[2]--;
      }
      for (int i = pos; i >= 1; i--) {
        counts[i]--;
      }
    }
    pos--;
  }
  cout << res << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
}

### Prompt

Your challenge is to write 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 factorial(int n) { return (n == 1 || n == 0) ? 1 : n * factorial(n - 1); }
void solve() {
  int digits;
  cin >> digits;
  string num;
  cin >> num;
  long long product = 1;
  string res = "";
  vector<int> counts(10, 0);
  for (int i = 0; i < digits; i++) {
    product *= factorial(num[i] - 48);
    for (int j = 1; j <= num[i] - 48; j++) {
      counts[j]++;
    }
  }
  for (int i = 1; i <= 9; i++) {
    if (i == 4) {
      while (counts[i] > 0) {
        counts[2] += 2;
        counts[i]--;
      }
    } else if (i == 6) {
      while (counts[i] > 0) {
        counts[2]++;
        counts[3]++;
        counts[i]--;
      }
    } else if (i == 8) {
      while (counts[i] > 0) {
        counts[2] += 3;
        counts[i]--;
      }
    } else if (i == 9) {
      while (counts[i] > 0) {
        counts[3] += 2;
        counts[i]--;
      }
    }
  }
  int pos = 9;
  while (pos != 1) {
    while (counts[pos] > 0) {
      res += char(pos + 48);
      if (pos == 7) {
        counts[3]--;
        counts[2]--;
        counts[2]--;
        counts[2]--;
      } else if (pos == 5) {
        counts[2]--;
        counts[2]--;
      }
      for (int i = pos; i >= 1; i--) {
        counts[i]--;
      }
    }
    pos--;
  }
  cout << res << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
}

```