#include <bits/stdc++.h>
using namespace std;
map<int, int> ma;
void set_mp() {
  ma[2] = 2;
  ma[3] = 3;
  ma[4] = 223;
  ma[5] = 5;
  ma[6] = 53;
  ma[7] = 7;
  ma[8] = 7222;
  ma[9] = 7332;
}
int main() {
  set_mp();
  int n;
  cin >> n;
  vector<int> v;
  string k;
  cin >> k;
  int a[20];
  for (int i = 0; i < k.length(); i++) a[i] = k[i] - 48;
  for (int i = 0; i < k.length(); i++) {
    if (a[i] > 1) {
      int k = ma[a[i]];
      while (k) {
        int l = k % 10;
        v.push_back(l);
        k /= 10;
      }
    }
  }
  sort(v.begin(), v.end());
  for (int i = v.size() - 1; i >= 0; i--) cout << v[i];
}

### 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;
map<int, int> ma;
void set_mp() {
  ma[2] = 2;
  ma[3] = 3;
  ma[4] = 223;
  ma[5] = 5;
  ma[6] = 53;
  ma[7] = 7;
  ma[8] = 7222;
  ma[9] = 7332;
}
int main() {
  set_mp();
  int n;
  cin >> n;
  vector<int> v;
  string k;
  cin >> k;
  int a[20];
  for (int i = 0; i < k.length(); i++) a[i] = k[i] - 48;
  for (int i = 0; i < k.length(); i++) {
    if (a[i] > 1) {
      int k = ma[a[i]];
      while (k) {
        int l = k % 10;
        v.push_back(l);
        k /= 10;
      }
    }
  }
  sort(v.begin(), v.end());
  for (int i = v.size() - 1; i >= 0; i--) cout << v[i];
}

```