#include <bits/stdc++.h>
using namespace std;
const int SIZE = 1000;
bool comp(const int a, const int b);
char RULE[][10] = {{"0"}, {"1"},  {"2"}, {"3"},    {"322"},
                   {"5"}, {"53"}, {"7"}, {"7222"}, {"7332"}};
int main(void) {
  int n;
  char s[SIZE];
  char ans[SIZE] = {'\0'};
  cin >> n;
  cin >> s;
  for (int i = 0; s[i]; i++)
    if (s[i] != '1' && s[i] != '0') strcat(ans, RULE[s[i] - '0']);
  sort(ans, ans + strlen(ans), comp);
  cout << ans << endl;
  return 0;
}
bool comp(const int a, const int b) { return a > b; }

### Prompt

Develop a solution in Cpp 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;
const int SIZE = 1000;
bool comp(const int a, const int b);
char RULE[][10] = {{"0"}, {"1"},  {"2"}, {"3"},    {"322"},
                   {"5"}, {"53"}, {"7"}, {"7222"}, {"7332"}};
int main(void) {
  int n;
  char s[SIZE];
  char ans[SIZE] = {'\0'};
  cin >> n;
  cin >> s;
  for (int i = 0; s[i]; i++)
    if (s[i] != '1' && s[i] != '0') strcat(ans, RULE[s[i] - '0']);
  sort(ans, ans + strlen(ans), comp);
  cout << ans << endl;
  return 0;
}
bool comp(const int a, const int b) { return a > b; }

```