#include <bits/stdc++.h>
using namespace std;
const int maxn = 10;
int n, a[maxn];
string str;
int main() {
  scanf("%d", &n);
  cin >> str;
  for (int i = 0; i < n; i++) {
    int d = str[i] - '0';
    if (d == 4)
      a[3]++, a[2] += 2;
    else if (d == 6)
      a[5]++, a[3]++;
    else if (d == 8)
      a[7]++, a[2] += 3;
    else if (d == 9)
      a[7]++, a[3] += 2, a[2]++;
    else
      a[d]++;
  }
  for (int d = 9; d > 1; d--)
    for (int i = 0; i < a[d]; i++) printf("%d", d);
  return 0;
}

### Prompt

Your challenge is to write 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;
const int maxn = 10;
int n, a[maxn];
string str;
int main() {
  scanf("%d", &n);
  cin >> str;
  for (int i = 0; i < n; i++) {
    int d = str[i] - '0';
    if (d == 4)
      a[3]++, a[2] += 2;
    else if (d == 6)
      a[5]++, a[3]++;
    else if (d == 8)
      a[7]++, a[2] += 3;
    else if (d == 9)
      a[7]++, a[3] += 2, a[2]++;
    else
      a[d]++;
  }
  for (int d = 9; d > 1; d--)
    for (int i = 0; i < a[d]; i++) printf("%d", d);
  return 0;
}

```