#include <bits/stdc++.h>
using namespace std;
int x = 10, y = 1, p, i, l, b[1000], n;
long long a;
int main() {
  cin >> n;
  cin >> a;
  for (i = 1; i <= n; i++) {
    p = (a % x - a % y) / y;
    a /= 10;
    if (p == 2) b[++l] = 2;
    if (p == 3) b[++l] = 3;
    if (p == 4) {
      b[++l] = 3;
      b[++l] = 2;
      b[++l] = 2;
    }
    if (p == 5) b[++l] = 5;
    if (p == 6) {
      b[++l] = 5;
      b[++l] = 3;
    }
    if (p == 7) b[++l] = 7;
    if (p == 8) {
      b[++l] = 7;
      b[++l] = 2;
      b[++l] = 2;
      b[++l] = 2;
    }
    if (p == 9) {
      b[++l] = 7;
      b[++l] = 3;
      b[++l] = 3;
      b[++l] = 2;
    }
  }
  sort(b + 1, b + 1 + l);
  for (i = l; i >= 1; i--) cout << b[i];
  cin >> i;
  return 0;
}

### Prompt

Your task is to create 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;
int x = 10, y = 1, p, i, l, b[1000], n;
long long a;
int main() {
  cin >> n;
  cin >> a;
  for (i = 1; i <= n; i++) {
    p = (a % x - a % y) / y;
    a /= 10;
    if (p == 2) b[++l] = 2;
    if (p == 3) b[++l] = 3;
    if (p == 4) {
      b[++l] = 3;
      b[++l] = 2;
      b[++l] = 2;
    }
    if (p == 5) b[++l] = 5;
    if (p == 6) {
      b[++l] = 5;
      b[++l] = 3;
    }
    if (p == 7) b[++l] = 7;
    if (p == 8) {
      b[++l] = 7;
      b[++l] = 2;
      b[++l] = 2;
      b[++l] = 2;
    }
    if (p == 9) {
      b[++l] = 7;
      b[++l] = 3;
      b[++l] = 3;
      b[++l] = 2;
    }
  }
  sort(b + 1, b + 1 + l);
  for (i = l; i >= 1; i--) cout << b[i];
  cin >> i;
  return 0;
}

```