#include <bits/stdc++.h>
using namespace std;
long long ten(int n) {
  if (n == 0)
    return 1;
  else if (n == 1)
    return 10;
  else
    return 10 * ten(n - 1);
}
int main() {
  int n, i;
  char as[100];
  cin >> n >> as;
  int k = 0, a[n + 2];
  int b[100000], p;
  k = 0;
  for (i = 0; i < n; i++) {
    if (as[i] == '2') {
      b[k] = 2;
      k = k + 1;
    } else if (as[i] == '3') {
      b[k] = 3;
      k = k + 1;
    } else if (as[i] == '5') {
      b[k] = 5;
      k = k + 1;
    } else if (as[i] == '7') {
      b[k] = 7;
      k = k + 1;
    } else if (as[i] == '9') {
      b[k] = 3;
      b[k + 1] = 3;
      b[k + 2] = 7;
      b[k + 3] = 2;
      k = k + 4;
    } else if (as[i] == '4') {
      b[k] = 2;
      b[k + 1] = 2;
      b[k + 2] = 3;
      k = k + 3;
    } else if (as[i] == '6') {
      b[k] = 3;
      b[k + 1] = 5;
      k = k + 2;
    } else if (as[i] == '8') {
      b[k] = 2;
      b[k + 1] = 2;
      b[k + 2] = 2;
      b[k + 3] = 7;
      k = k + 4;
      ;
    } else
      continue;
  }
  sort(b, b + k);
  for (i = k - 1; i >= 0; i--) cout << b[i];
  cout << endl;
}

### 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;
long long ten(int n) {
  if (n == 0)
    return 1;
  else if (n == 1)
    return 10;
  else
    return 10 * ten(n - 1);
}
int main() {
  int n, i;
  char as[100];
  cin >> n >> as;
  int k = 0, a[n + 2];
  int b[100000], p;
  k = 0;
  for (i = 0; i < n; i++) {
    if (as[i] == '2') {
      b[k] = 2;
      k = k + 1;
    } else if (as[i] == '3') {
      b[k] = 3;
      k = k + 1;
    } else if (as[i] == '5') {
      b[k] = 5;
      k = k + 1;
    } else if (as[i] == '7') {
      b[k] = 7;
      k = k + 1;
    } else if (as[i] == '9') {
      b[k] = 3;
      b[k + 1] = 3;
      b[k + 2] = 7;
      b[k + 3] = 2;
      k = k + 4;
    } else if (as[i] == '4') {
      b[k] = 2;
      b[k + 1] = 2;
      b[k + 2] = 3;
      k = k + 3;
    } else if (as[i] == '6') {
      b[k] = 3;
      b[k + 1] = 5;
      k = k + 2;
    } else if (as[i] == '8') {
      b[k] = 2;
      b[k + 1] = 2;
      b[k + 2] = 2;
      b[k + 3] = 7;
      k = k + 4;
      ;
    } else
      continue;
  }
  sort(b, b + k);
  for (i = k - 1; i >= 0; i--) cout << b[i];
  cout << endl;
}

```