#include <bits/stdc++.h>
using namespace std;
int f[5];
void add(int n) {
  switch (n) {
    case 2:
      ++f[0];
      break;
    case 3:
      ++f[1];
      break;
    case 4:
      ++f[1];
      f[0] += 2;
      break;
    case 5:
      ++f[2];
      break;
    case 6:
      ++f[2];
      ++f[1];
      break;
    case 7:
      ++f[3];
      break;
    case 8:
      ++f[3];
      f[0] += 3;
      break;
    case 9:
      ++f[3];
      ++f[0];
      f[1] += 2;
      break;
  }
}
int main() {
  int n, i, j, c = 0;
  char a[20], op[100];
  scanf("%d", &n);
  scanf("%s", &a);
  for (i = 0; i < 4; ++i) f[i] = 0;
  for (i = 0; i < n; ++i) {
    a[i] = a[i] - '0';
  }
  for (i = 0; i < n; ++i) {
    add(a[i]);
  }
  for (i = 1; i <= f[3]; ++i) op[c++] = '7';
  for (i = 1; i <= f[2]; ++i) op[c++] = '5';
  for (i = 1; i <= f[1]; ++i) op[c++] = '3';
  for (i = 1; i <= f[0]; ++i) op[c++] = '2';
  for (i = 0; i < c; ++i) printf("%c", op[i]);
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 f[5];
void add(int n) {
  switch (n) {
    case 2:
      ++f[0];
      break;
    case 3:
      ++f[1];
      break;
    case 4:
      ++f[1];
      f[0] += 2;
      break;
    case 5:
      ++f[2];
      break;
    case 6:
      ++f[2];
      ++f[1];
      break;
    case 7:
      ++f[3];
      break;
    case 8:
      ++f[3];
      f[0] += 3;
      break;
    case 9:
      ++f[3];
      ++f[0];
      f[1] += 2;
      break;
  }
}
int main() {
  int n, i, j, c = 0;
  char a[20], op[100];
  scanf("%d", &n);
  scanf("%s", &a);
  for (i = 0; i < 4; ++i) f[i] = 0;
  for (i = 0; i < n; ++i) {
    a[i] = a[i] - '0';
  }
  for (i = 0; i < n; ++i) {
    add(a[i]);
  }
  for (i = 1; i <= f[3]; ++i) op[c++] = '7';
  for (i = 1; i <= f[2]; ++i) op[c++] = '5';
  for (i = 1; i <= f[1]; ++i) op[c++] = '3';
  for (i = 1; i <= f[0]; ++i) op[c++] = '2';
  for (i = 0; i < c; ++i) printf("%c", op[i]);
}

```