#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, s = 1;
  char a;
  string b[10], c;
  cin >> n;
  b[0] = "";
  b[1] = "";
  b[2] = "2";
  b[3] = "3";
  b[4] = "322";
  b[5] = "5";
  b[6] = "53";
  b[7] = "7";
  b[8] = "7222";
  b[9] = "7332";
  int k = 0;
  int d = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a;
    s = a - 48;
    if (s != 1 && s != 0) c = c + b[s];
  }
  sort(c.begin(), c.end());
  k = c.size();
  for (int i = k - 1; i >= 0; i--) cout << c[i];
  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() {
  long long n, s = 1;
  char a;
  string b[10], c;
  cin >> n;
  b[0] = "";
  b[1] = "";
  b[2] = "2";
  b[3] = "3";
  b[4] = "322";
  b[5] = "5";
  b[6] = "53";
  b[7] = "7";
  b[8] = "7222";
  b[9] = "7332";
  int k = 0;
  int d = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a;
    s = a - 48;
    if (s != 1 && s != 0) c = c + b[s];
  }
  sort(c.begin(), c.end());
  k = c.size();
  for (int i = k - 1; i >= 0; i--) cout << c[i];
  return 0;
}

```