#include <bits/stdc++.h>
using namespace std;
const int N = 100;
int main() {
  int n;
  int a;
  int v[10];
  while (~scanf("%d", &n)) {
    memset(v, 0, sizeof(v));
    for (int i = 0; i < n; i++) {
      scanf("%1d", &a);
      switch (a) {
        case 2:
          v[2]++;
          break;
        case 3:
          v[3]++;
          break;
        case 4:
          v[3]++;
          v[2] += 2;
          break;
        case 5:
          v[5]++;
          break;
        case 6:
          v[5]++;
          v[3]++;
          break;
        case 7:
          v[7]++;
          break;
        case 8:
          v[7]++;
          v[2] += 3;
          break;
        case 9:
          v[7]++;
          v[3] += 2;
          v[2]++;
          break;
        default:
          break;
      }
    }
    for (int i = 9; i >= 0; i--) {
      for (int j = 0; j < v[i]; j++) {
        printf("%d", i);
      }
    }
    printf("\n");
  }
  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;
const int N = 100;
int main() {
  int n;
  int a;
  int v[10];
  while (~scanf("%d", &n)) {
    memset(v, 0, sizeof(v));
    for (int i = 0; i < n; i++) {
      scanf("%1d", &a);
      switch (a) {
        case 2:
          v[2]++;
          break;
        case 3:
          v[3]++;
          break;
        case 4:
          v[3]++;
          v[2] += 2;
          break;
        case 5:
          v[5]++;
          break;
        case 6:
          v[5]++;
          v[3]++;
          break;
        case 7:
          v[7]++;
          break;
        case 8:
          v[7]++;
          v[2] += 3;
          break;
        case 9:
          v[7]++;
          v[3] += 2;
          v[2]++;
          break;
        default:
          break;
      }
    }
    for (int i = 9; i >= 0; i--) {
      for (int j = 0; j < v[i]; j++) {
        printf("%d", i);
      }
    }
    printf("\n");
  }
  return 0;
}

```