#include <bits/stdc++.h>
using namespace std;
int ans[100];
int cou = 0;
char s[20];
int main() {
  int n;
  int t;
  scanf("%d %s", &n, s);
  for (int i = 0; i < n; ++i) {
    t = s[i] - '0';
    if (t <= 1) continue;
    if ((t & 1 && t != 9) || t == 2) ans[cou++] = t;
    if (t == 4) {
      ans[cou++] = 3;
      ans[cou++] = 2;
      ans[cou++] = 2;
    }
    if (t == 6) {
      ans[cou++] = 5;
      ans[cou++] = 3;
    }
    if (t == 8) {
      ans[cou++] = 7;
      ans[cou++] = 2;
      ans[cou++] = 2;
      ans[cou++] = 2;
    }
    if (t == 9) {
      ans[cou++] = 7;
      ans[cou++] = 2;
      ans[cou++] = 3;
      ans[cou++] = 3;
    }
  }
  sort(ans, ans + cou);
  for (int i = cou - 1; i >= 0; --i) cout << ans[i];
  cout << endl;
  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 ans[100];
int cou = 0;
char s[20];
int main() {
  int n;
  int t;
  scanf("%d %s", &n, s);
  for (int i = 0; i < n; ++i) {
    t = s[i] - '0';
    if (t <= 1) continue;
    if ((t & 1 && t != 9) || t == 2) ans[cou++] = t;
    if (t == 4) {
      ans[cou++] = 3;
      ans[cou++] = 2;
      ans[cou++] = 2;
    }
    if (t == 6) {
      ans[cou++] = 5;
      ans[cou++] = 3;
    }
    if (t == 8) {
      ans[cou++] = 7;
      ans[cou++] = 2;
      ans[cou++] = 2;
      ans[cou++] = 2;
    }
    if (t == 9) {
      ans[cou++] = 7;
      ans[cou++] = 2;
      ans[cou++] = 3;
      ans[cou++] = 3;
    }
  }
  sort(ans, ans + cou);
  for (int i = cou - 1; i >= 0; --i) cout << ans[i];
  cout << endl;
  return 0;
}

```