#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  int s[100];
  char c[100];
  while (cin >> n) {
    memset(s, 0, sizeof(s));
    memset(c, 0, sizeof(c));
    cin >> c;
    for (int i = 0; c[i]; i++) {
      if (c[i] == '2') s[2]++;
      if (c[i] == '3') s[3]++;
      if (c[i] == '4') s[2] += 2, s[3]++;
      if (c[i] == '5') s[5]++;
      if (c[i] == '6') s[3]++, s[5]++;
      if (c[i] == '7') s[7]++;
      if (c[i] == '8') s[2] += 3, s[7]++;
      if (c[i] == '9') s[3] += 2, s[2]++, s[7]++;
    }
    for (int i = 9; i > 1; i--) {
      if (s[i]) {
        for (int j = 0; j < s[i]; j++) printf("%d", i);
      }
    }
    printf("\n");
  }
}

### 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;
int main() {
  int n;
  int s[100];
  char c[100];
  while (cin >> n) {
    memset(s, 0, sizeof(s));
    memset(c, 0, sizeof(c));
    cin >> c;
    for (int i = 0; c[i]; i++) {
      if (c[i] == '2') s[2]++;
      if (c[i] == '3') s[3]++;
      if (c[i] == '4') s[2] += 2, s[3]++;
      if (c[i] == '5') s[5]++;
      if (c[i] == '6') s[3]++, s[5]++;
      if (c[i] == '7') s[7]++;
      if (c[i] == '8') s[2] += 3, s[7]++;
      if (c[i] == '9') s[3] += 2, s[2]++, s[7]++;
    }
    for (int i = 9; i > 1; i--) {
      if (s[i]) {
        for (int j = 0; j < s[i]; j++) printf("%d", i);
      }
    }
    printf("\n");
  }
}

```