#include <bits/stdc++.h>
using namespace std;
int main() {
  int a[10];
  int t, n, i;
  for (i = 0; i < 10; ++i) {
    a[i] = 0;
  }
  scanf("%d", &n);
  getchar();
  vector<int> u;
  for (i = 0; i < n; ++i) {
    t = getchar() - '0';
    switch (t) {
      case (2):
        a[2] += 1;
        break;
      case (3):
        a[3] += 1;
        a[2] += 1;
        break;
      case (4):
        a[2] += 3;
        a[3] += 1;
        break;
      case (5):
        u.push_back(5);
        break;
      case (6):
        u.push_back(5);
        a[2] += 1;
        a[3] += 1;
        break;
      case (7):
        u.push_back(7);
        break;
      case (8):
        u.push_back(7);
        a[2] += 3;
        break;
      case (9):
        u.push_back(7);
        a[2] += 3;
        a[3] += 2;
        break;
    }
  }
  a[2] -= a[3];
  for (i = 0; i < a[3]; ++i) {
    u.push_back(3);
  }
  for (i = 0; i < a[2]; ++i) {
    u.push_back(2);
  }
  sort(u.begin(), u.end());
  for (i = u.size() - 1; i > -1; --i) {
    printf("%d", u[i]);
  }
  return 0;
}

### Prompt

Your challenge is to write 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 main() {
  int a[10];
  int t, n, i;
  for (i = 0; i < 10; ++i) {
    a[i] = 0;
  }
  scanf("%d", &n);
  getchar();
  vector<int> u;
  for (i = 0; i < n; ++i) {
    t = getchar() - '0';
    switch (t) {
      case (2):
        a[2] += 1;
        break;
      case (3):
        a[3] += 1;
        a[2] += 1;
        break;
      case (4):
        a[2] += 3;
        a[3] += 1;
        break;
      case (5):
        u.push_back(5);
        break;
      case (6):
        u.push_back(5);
        a[2] += 1;
        a[3] += 1;
        break;
      case (7):
        u.push_back(7);
        break;
      case (8):
        u.push_back(7);
        a[2] += 3;
        break;
      case (9):
        u.push_back(7);
        a[2] += 3;
        a[3] += 2;
        break;
    }
  }
  a[2] -= a[3];
  for (i = 0; i < a[3]; ++i) {
    u.push_back(3);
  }
  for (i = 0; i < a[2]; ++i) {
    u.push_back(2);
  }
  sort(u.begin(), u.end());
  for (i = u.size() - 1; i > -1; --i) {
    printf("%d", u[i]);
  }
  return 0;
}

```