#include <bits/stdc++.h>
using namespace std;
void print() { cout << "" << endl; }
template <typename T, typename... Types>
void print(T var1, Types... var2) {
  cout << var1 << " ";
  print(var2...);
}
const int mode = 1000000007;
const double PI = acos(-1);
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int i, j, k, x, y, sum = 0, cnt = 0, flag = 0, n, m;
  cin >> n;
  string s;
  cin >> s;
  vector<long long int> v;
  int a[10] = {0, 1, 2, 3, 322, 5, 53, 7, 7222, 7332};
  for (i = 0; s[i]; i++) {
    if (s[i] == '1' || s[i] == '0')
      ;
    else {
      k = (a[(int)(s[i] - '0')]);
      while (k != 0) {
        v.push_back(k % 10);
        k /= 10;
      }
    }
  }
  sort((v).begin(), (v).end());
  reverse((v).begin(), (v).end());
  for (i = 0; i < v.size(); i++) {
    cout << v[i];
  }
  return 0;
}

### Prompt

Generate 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 print() { cout << "" << endl; }
template <typename T, typename... Types>
void print(T var1, Types... var2) {
  cout << var1 << " ";
  print(var2...);
}
const int mode = 1000000007;
const double PI = acos(-1);
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int i, j, k, x, y, sum = 0, cnt = 0, flag = 0, n, m;
  cin >> n;
  string s;
  cin >> s;
  vector<long long int> v;
  int a[10] = {0, 1, 2, 3, 322, 5, 53, 7, 7222, 7332};
  for (i = 0; s[i]; i++) {
    if (s[i] == '1' || s[i] == '0')
      ;
    else {
      k = (a[(int)(s[i] - '0')]);
      while (k != 0) {
        v.push_back(k % 10);
        k /= 10;
      }
    }
  }
  sort((v).begin(), (v).end());
  reverse((v).begin(), (v).end());
  for (i = 0; i < v.size(); i++) {
    cout << v[i];
  }
  return 0;
}

```