#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, a[10][10] = {0}, ans[10] = {0}, i, j, x;
  string s;
  a[2][2] = 1;
  a[3][3] = 1;
  a[5][5] = 1;
  a[7][7] = 1;
  a[4][3] = 1;
  a[4][2] = 2;
  a[6][5] = 1;
  a[6][3] = 1;
  a[8][7] = 1;
  a[8][2] = 3;
  a[9][7] = 1;
  a[9][3] = 2;
  a[9][2] = 1;
  cin >> n >> s;
  for (i = 0; i <= n - 1; i++) {
    if (s[i] != '0' && s[i] != '1') {
      x = s[i] - '0';
      for (j = 2; j <= 9; j++) {
        ans[j] += a[x][j];
      }
    }
  }
  for (i = 9; i >= 2; i--) {
    for (j = ans[i]; j >= 1; j--) {
      cout << i;
    }
  }
  return 0;
}

### Prompt

Your task is to create 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 n, a[10][10] = {0}, ans[10] = {0}, i, j, x;
  string s;
  a[2][2] = 1;
  a[3][3] = 1;
  a[5][5] = 1;
  a[7][7] = 1;
  a[4][3] = 1;
  a[4][2] = 2;
  a[6][5] = 1;
  a[6][3] = 1;
  a[8][7] = 1;
  a[8][2] = 3;
  a[9][7] = 1;
  a[9][3] = 2;
  a[9][2] = 1;
  cin >> n >> s;
  for (i = 0; i <= n - 1; i++) {
    if (s[i] != '0' && s[i] != '1') {
      x = s[i] - '0';
      for (j = 2; j <= 9; j++) {
        ans[j] += a[x][j];
      }
    }
  }
  for (i = 9; i >= 2; i--) {
    for (j = ans[i]; j >= 1; j--) {
      cout << i;
    }
  }
  return 0;
}

```