#include <bits/stdc++.h>
using namespace std;
const int N = 1e5;
long long a[N], b[N], j = 1;
void fun(long long n) {
  int i, k, m;
  for (i = n; i > 1; i--) {
    m = i;
    if (m % 2 != 0 && m != 9) {
      b[j++] = m;
      break;
    }
    if (m == 9) {
      b[j++] = 7;
      b[j++] = 3;
      b[j++] = 3;
      b[j++] = 2;
      break;
    }
    if (m == 6) {
      b[j++] = 5;
      b[j++] = 3;
      break;
    }
    for (k = 2; k <= m; k++) {
      if (m % k == 0) {
        b[j++] = k;
        m = m / k;
        k = 1;
      }
    }
  }
}
int main() {
  long long n, i, k, p, y = 0, sum = 0;
  char ch;
  cin >> n;
  getchar();
  for (i = 0; i < n; i++) {
    cin >> ch;
    a[i] = ch - '0';
    fun(a[i]);
  }
  sort(b + 1, b + j);
  for (i = j - 1; i >= 1; i--) cout << b[i];
  cout << endl;
  return 0;
}

### Prompt

Generate 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 N = 1e5;
long long a[N], b[N], j = 1;
void fun(long long n) {
  int i, k, m;
  for (i = n; i > 1; i--) {
    m = i;
    if (m % 2 != 0 && m != 9) {
      b[j++] = m;
      break;
    }
    if (m == 9) {
      b[j++] = 7;
      b[j++] = 3;
      b[j++] = 3;
      b[j++] = 2;
      break;
    }
    if (m == 6) {
      b[j++] = 5;
      b[j++] = 3;
      break;
    }
    for (k = 2; k <= m; k++) {
      if (m % k == 0) {
        b[j++] = k;
        m = m / k;
        k = 1;
      }
    }
  }
}
int main() {
  long long n, i, k, p, y = 0, sum = 0;
  char ch;
  cin >> n;
  getchar();
  for (i = 0; i < n; i++) {
    cin >> ch;
    a[i] = ch - '0';
    fun(a[i]);
  }
  sort(b + 1, b + j);
  for (i = j - 1; i >= 1; i--) cout << b[i];
  cout << endl;
  return 0;
}

```