#include <bits/stdc++.h>
using namespace std;
long long fact(int a) {
  if ((a == 0) || (a == 1))
    return (1);
  else
    return (a * fact(a - 1));
}
int main() {
  vector<int> t;
  vector<int> t1;
  long long int k = 0, a, j, i, n;
  cin >> n;
  cin >> a;
  t.resize(n);
  for (i = 0; i < n; i++) {
    t[i] = a % 10;
    a = a / 10;
  }
  for (i = 0; i < n; i++) {
    if (t[i] == 4) {
      t1.push_back(3);
      t1.push_back(2);
      t1.push_back(2);
      k += 2;
    } else if (t[i] == 8) {
      t1.push_back(7);
      t1.push_back(2);
      t1.push_back(2);
      t1.push_back(2);
      k += 3;
    } else if (t[i] == 6) {
      t1.push_back(5);
      t1.push_back(3);
      k += 1;
    } else if (t[i] == 9) {
      t1.push_back(7);
      t1.push_back(3);
      t1.push_back(2);
      t1.push_back(3);
      k += 3;
    } else if (t[i] < 2)
      k -= 1;
    else
      t1.push_back(t[i]);
    k += 1;
  }
  n = k;
  t1.resize(n);
  stable_sort(t1.begin(), t1.end());
  reverse(t1.begin(), t1.end());
  for (i = 0; i < n; i++) cout << t1[i];
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 fact(int a) {
  if ((a == 0) || (a == 1))
    return (1);
  else
    return (a * fact(a - 1));
}
int main() {
  vector<int> t;
  vector<int> t1;
  long long int k = 0, a, j, i, n;
  cin >> n;
  cin >> a;
  t.resize(n);
  for (i = 0; i < n; i++) {
    t[i] = a % 10;
    a = a / 10;
  }
  for (i = 0; i < n; i++) {
    if (t[i] == 4) {
      t1.push_back(3);
      t1.push_back(2);
      t1.push_back(2);
      k += 2;
    } else if (t[i] == 8) {
      t1.push_back(7);
      t1.push_back(2);
      t1.push_back(2);
      t1.push_back(2);
      k += 3;
    } else if (t[i] == 6) {
      t1.push_back(5);
      t1.push_back(3);
      k += 1;
    } else if (t[i] == 9) {
      t1.push_back(7);
      t1.push_back(3);
      t1.push_back(2);
      t1.push_back(3);
      k += 3;
    } else if (t[i] < 2)
      k -= 1;
    else
      t1.push_back(t[i]);
    k += 1;
  }
  n = k;
  t1.resize(n);
  stable_sort(t1.begin(), t1.end());
  reverse(t1.begin(), t1.end());
  for (i = 0; i < n; i++) cout << t1[i];
  return 0;
}

```