#include <bits/stdc++.h>
using namespace std;
string s;
int n, cnt[15];
void init(int num) {
  if (num >= 2) cnt[2]++;
  if (num >= 3) cnt[3]++;
  if (num >= 4) cnt[2] += 2;
  if (num >= 5) cnt[5]++;
  if (num >= 6) cnt[2]++, cnt[3]++;
  if (num >= 7) cnt[7]++;
  if (num >= 8) cnt[2] += 3;
  if (num >= 9) cnt[3] += 2;
}
int main() {
  cin >> n >> s;
  for (int i = 0; i < n; i++) {
    int num = s[i] - '0';
    init(num);
  }
  for (int i = 1; i <= cnt[7]; i++)
    printf("7"), cnt[2] -= 4, cnt[3] -= 2, cnt[5]--;
  for (int i = 1; i <= cnt[5]; i++) printf("5"), cnt[2] -= 3, cnt[3]--;
  for (int i = 1; i <= cnt[3]; i++) printf("3"), cnt[2]--;
  for (int i = 1; i <= cnt[2]; i++) printf("2");
  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;
string s;
int n, cnt[15];
void init(int num) {
  if (num >= 2) cnt[2]++;
  if (num >= 3) cnt[3]++;
  if (num >= 4) cnt[2] += 2;
  if (num >= 5) cnt[5]++;
  if (num >= 6) cnt[2]++, cnt[3]++;
  if (num >= 7) cnt[7]++;
  if (num >= 8) cnt[2] += 3;
  if (num >= 9) cnt[3] += 2;
}
int main() {
  cin >> n >> s;
  for (int i = 0; i < n; i++) {
    int num = s[i] - '0';
    init(num);
  }
  for (int i = 1; i <= cnt[7]; i++)
    printf("7"), cnt[2] -= 4, cnt[3] -= 2, cnt[5]--;
  for (int i = 1; i <= cnt[5]; i++) printf("5"), cnt[2] -= 3, cnt[3]--;
  for (int i = 1; i <= cnt[3]; i++) printf("3"), cnt[2]--;
  for (int i = 1; i <= cnt[2]; i++) printf("2");
  return 0;
}

```