#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  long long tt;
  while (cin >> n) {
    cin >> tt;
    int a[10] = {0};
    while (tt) {
      int t = tt % 10;
      tt /= 10;
      if (t == 2) {
        a[2]++;
      } else if (t == 3) {
        a[3]++;
      } else if (t == 4) {
        a[3]++;
        a[2] += 2;
      } else if (t == 5) {
        a[5]++;
      } else if (t == 6) {
        a[5]++;
        a[3]++;
      } else if (t == 7) {
        a[7]++;
      } else if (t == 8) {
        a[7]++;
        a[2] += 3;
      } else if (t == 9) {
        a[7]++;
        a[3] += 2;
        a[2]++;
      }
    }
    for (int i = 9; i > 1; i--) {
      for (int j = 0; j < a[i]; j++) {
        cout << i;
      }
    }
    cout << endl;
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 n;
  long long tt;
  while (cin >> n) {
    cin >> tt;
    int a[10] = {0};
    while (tt) {
      int t = tt % 10;
      tt /= 10;
      if (t == 2) {
        a[2]++;
      } else if (t == 3) {
        a[3]++;
      } else if (t == 4) {
        a[3]++;
        a[2] += 2;
      } else if (t == 5) {
        a[5]++;
      } else if (t == 6) {
        a[5]++;
        a[3]++;
      } else if (t == 7) {
        a[7]++;
      } else if (t == 8) {
        a[7]++;
        a[2] += 3;
      } else if (t == 9) {
        a[7]++;
        a[3] += 2;
        a[2]++;
      }
    }
    for (int i = 9; i > 1; i--) {
      for (int j = 0; j < a[i]; j++) {
        cout << i;
      }
    }
    cout << endl;
  }
  return 0;
}

```