#include <bits/stdc++.h>
using namespace std;
int v[11], i, a, n;
char w[16], x;
int main() {
  cin >> n;
  cin >> w;
  for (i = 0; i < n; i++) {
    x = w[i];
    x -= '0';
    if (x == 4) {
      v[2] += 2;
      v[3]++;
    }
    if (x == 6) {
      v[3]++;
      v[5]++;
    }
    if (x == 8) {
      v[7]++;
      v[2] += 3;
    }
    if (x == 9) {
      v[7]++;
      v[2]++;
      v[3] += 2;
    }
    if (x == 2 || x == 3 || x == 5 || x == 7) v[x]++;
    a /= 10;
  }
  for (i = 7; i > 1; i--)
    if (v[i]) {
      while (v[i]) {
        cout << i;
        v[i]--;
      }
    }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 v[11], i, a, n;
char w[16], x;
int main() {
  cin >> n;
  cin >> w;
  for (i = 0; i < n; i++) {
    x = w[i];
    x -= '0';
    if (x == 4) {
      v[2] += 2;
      v[3]++;
    }
    if (x == 6) {
      v[3]++;
      v[5]++;
    }
    if (x == 8) {
      v[7]++;
      v[2] += 3;
    }
    if (x == 9) {
      v[7]++;
      v[2]++;
      v[3] += 2;
    }
    if (x == 2 || x == 3 || x == 5 || x == 7) v[x]++;
    a /= 10;
  }
  for (i = 7; i > 1; i--)
    if (v[i]) {
      while (v[i]) {
        cout << i;
        v[i]--;
      }
    }
  return 0;
}

```