#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 7;
int n, cnt[10];
char s[N];
int main() {
  scanf("%d", &n);
  scanf("%s", s + 1);
  for (int i = (1), iE = (n); i <= iE; i++) {
    if (s[i] == '1')
      ;
    else if (s[i] == '2')
      cnt[2]++;
    else if (s[i] == '3')
      cnt[3]++;
    else if (s[i] == '4')
      cnt[2] += 2, cnt[3]++;
    else if (s[i] == '5')
      cnt[5]++;
    else if (s[i] == '6')
      cnt[5]++, cnt[3]++;
    else if (s[i] == '7')
      cnt[7]++;
    else if (s[i] == '8')
      cnt[2] += 3, cnt[7]++;
    else if (s[i] == '9')
      cnt[7]++, cnt[2]++, cnt[3] += 2;
  }
  for (int i = (9), iE = (1); i >= iE; i--)
    for (int j = (1), jE = (cnt[i]); j <= jE; j++) printf("%d", i);
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
const int N = 1e5 + 7;
int n, cnt[10];
char s[N];
int main() {
  scanf("%d", &n);
  scanf("%s", s + 1);
  for (int i = (1), iE = (n); i <= iE; i++) {
    if (s[i] == '1')
      ;
    else if (s[i] == '2')
      cnt[2]++;
    else if (s[i] == '3')
      cnt[3]++;
    else if (s[i] == '4')
      cnt[2] += 2, cnt[3]++;
    else if (s[i] == '5')
      cnt[5]++;
    else if (s[i] == '6')
      cnt[5]++, cnt[3]++;
    else if (s[i] == '7')
      cnt[7]++;
    else if (s[i] == '8')
      cnt[2] += 3, cnt[7]++;
    else if (s[i] == '9')
      cnt[7]++, cnt[2]++, cnt[3] += 2;
  }
  for (int i = (9), iE = (1); i >= iE; i--)
    for (int j = (1), jE = (cnt[i]); j <= jE; j++) printf("%d", i);
  return 0;
}

```