#include <bits/stdc++.h>
using namespace std;
int a[10] = {0}, b[1000], c = 0, i, j, n, m;
void p(int ch) {
  if (ch == 4)
    b[c++] = 2, b[c++] = 2, b[c++] = 3;
  else if (ch == 6)
    b[c++] = 5, b[c++] = 3;
  else if (ch == 8)
    b[c++] = 7, b[c++] = 2, b[c++] = 2, b[c++] = 2;
  else if (ch == 9)
    b[c++] = 7, b[c++] = 3, b[c++] = 3, b[c++] = 2;
  else
    b[c++] = ch;
}
int main() {
  string s;
  cin >> n >> s;
  for (i = 0; i < n; i++)
    if (s[i] - 48 >= 2) p(s[i] - 48);
  sort(b, b + c);
  for (i = c - 1; i >= 0; i--) cout << b[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 a[10] = {0}, b[1000], c = 0, i, j, n, m;
void p(int ch) {
  if (ch == 4)
    b[c++] = 2, b[c++] = 2, b[c++] = 3;
  else if (ch == 6)
    b[c++] = 5, b[c++] = 3;
  else if (ch == 8)
    b[c++] = 7, b[c++] = 2, b[c++] = 2, b[c++] = 2;
  else if (ch == 9)
    b[c++] = 7, b[c++] = 3, b[c++] = 3, b[c++] = 2;
  else
    b[c++] = ch;
}
int main() {
  string s;
  cin >> n >> s;
  for (i = 0; i < n; i++)
    if (s[i] - 48 >= 2) p(s[i] - 48);
  sort(b, b + c);
  for (i = c - 1; i >= 0; i--) cout << b[i];
  cout << endl;
  return 0;
}

```