#include <bits/stdc++.h>
using namespace std;
int char_to_int(char a) { return (int)a - 48; }
int main() {
  int pole[10];
  for (int i = 0; i < 10; i++) pole[i] = 0;
  int a;
  string j;
  cin >> a >> j;
  for (int i = 0; i < j.size(); i++) {
    int g;
    g = char_to_int(j[i]);
    for (int k = 2; k <= g; k++) pole[k]++;
  }
  pole[3] += 2 * pole[9];
  pole[2] += 3 * pole[8];
  for (int i = 0; i < pole[7]; i++) cout << 7;
  for (int i = 2; i < 7; i++) pole[i] -= pole[7];
  pole[2] += pole[6];
  pole[3] += pole[6];
  for (int i = 0; i < pole[5]; i++) cout << 5;
  for (int i = 2; i < 5; i++) pole[i] -= pole[5];
  pole[2] += 2 * pole[4];
  for (int i = 0; i < pole[3]; i++) cout << 3;
  pole[2] -= pole[3];
  for (int i = 0; i < pole[2]; i++) cout << 2;
  cout << endl;
  return 0;
}

### Prompt

Please formulate 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 char_to_int(char a) { return (int)a - 48; }
int main() {
  int pole[10];
  for (int i = 0; i < 10; i++) pole[i] = 0;
  int a;
  string j;
  cin >> a >> j;
  for (int i = 0; i < j.size(); i++) {
    int g;
    g = char_to_int(j[i]);
    for (int k = 2; k <= g; k++) pole[k]++;
  }
  pole[3] += 2 * pole[9];
  pole[2] += 3 * pole[8];
  for (int i = 0; i < pole[7]; i++) cout << 7;
  for (int i = 2; i < 7; i++) pole[i] -= pole[7];
  pole[2] += pole[6];
  pole[3] += pole[6];
  for (int i = 0; i < pole[5]; i++) cout << 5;
  for (int i = 2; i < 5; i++) pole[i] -= pole[5];
  pole[2] += 2 * pole[4];
  for (int i = 0; i < pole[3]; i++) cout << 3;
  pole[2] -= pole[3];
  for (int i = 0; i < pole[2]; i++) cout << 2;
  cout << endl;
  return 0;
}

```