#include <bits/stdc++.h>
using namespace std;
int n;
string s;
int c[10];
int main() {
  cin >> n;
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '0') continue;
    if (s[i] == '1') continue;
    if (s[i] >= '2') c[2]++;
    if (s[i] >= '3') c[3]++;
    if (s[i] >= '4') c[2] += 2;
    if (s[i] >= '5') c[5]++;
    if (s[i] >= '6') {
      c[2]++;
      c[3]++;
    }
    if (s[i] >= '7') c[7]++;
    if (s[i] >= '8') c[2] += 3;
    if (s[i] >= '9') c[3] += 2;
  }
  while (c[7]) {
    cout << 7;
    c[7]--;
    c[5]--;
    c[3] -= 2;
    c[2] -= 4;
  }
  while (c[5]) {
    cout << 5;
    c[5]--;
    c[3]--;
    c[2] -= 3;
  }
  while (c[3]) {
    cout << 3;
    c[3]--;
    c[2]--;
  }
  while (c[2]) {
    cout << 2;
    c[2]--;
  }
  cout << "\n";
}

### 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;
int n;
string s;
int c[10];
int main() {
  cin >> n;
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '0') continue;
    if (s[i] == '1') continue;
    if (s[i] >= '2') c[2]++;
    if (s[i] >= '3') c[3]++;
    if (s[i] >= '4') c[2] += 2;
    if (s[i] >= '5') c[5]++;
    if (s[i] >= '6') {
      c[2]++;
      c[3]++;
    }
    if (s[i] >= '7') c[7]++;
    if (s[i] >= '8') c[2] += 3;
    if (s[i] >= '9') c[3] += 2;
  }
  while (c[7]) {
    cout << 7;
    c[7]--;
    c[5]--;
    c[3] -= 2;
    c[2] -= 4;
  }
  while (c[5]) {
    cout << 5;
    c[5]--;
    c[3]--;
    c[2] -= 3;
  }
  while (c[3]) {
    cout << 3;
    c[3]--;
    c[2]--;
  }
  while (c[2]) {
    cout << 2;
    c[2]--;
  }
  cout << "\n";
}

```