#include <bits/stdc++.h>
using namespace std;
int n, tot, a[1010];
char ch;
int main() {
  scanf("%d\n", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%c", &ch);
    if (ch <= '1') continue;
    if (ch == '2') a[++tot] = ch - '0';
    if (ch == '3') a[++tot] = ch - '0';
    if (ch == '4') a[++tot] = 3, a[++tot] = 2, a[++tot] = 2;
    if (ch == '5') a[++tot] = 5;
    if (ch == '6') a[++tot] = 5, a[++tot] = 3;
    if (ch == '7') a[++tot] = 7;
    if (ch == '8') a[++tot] = 7, a[++tot] = 2, a[++tot] = 2, a[++tot] = 2;
    if (ch == '9') a[++tot] = 7, a[++tot] = 2, a[++tot] = 3, a[++tot] = 3;
  }
  sort(a + 1, a + 1 + tot);
  for (int i = tot; i; i--) putchar(a[i] + '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;
int n, tot, a[1010];
char ch;
int main() {
  scanf("%d\n", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%c", &ch);
    if (ch <= '1') continue;
    if (ch == '2') a[++tot] = ch - '0';
    if (ch == '3') a[++tot] = ch - '0';
    if (ch == '4') a[++tot] = 3, a[++tot] = 2, a[++tot] = 2;
    if (ch == '5') a[++tot] = 5;
    if (ch == '6') a[++tot] = 5, a[++tot] = 3;
    if (ch == '7') a[++tot] = 7;
    if (ch == '8') a[++tot] = 7, a[++tot] = 2, a[++tot] = 2, a[++tot] = 2;
    if (ch == '9') a[++tot] = 7, a[++tot] = 2, a[++tot] = 3, a[++tot] = 3;
  }
  sort(a + 1, a + 1 + tot);
  for (int i = tot; i; i--) putchar(a[i] + '0');
}

```