#include <bits/stdc++.h>
using namespace std;
const int mxn = 2e5 + 5;
void solve() {
  long long int n, k, m;
  long long int w;
  cin >> n;
  string s;
  cin >> s;
  long long int mark[12] = {0};
  for (long long int i = 0; i < n; i++) {
    int j = s[i] - '0';
    if (j == 1 || j == 0) continue;
    if (j == 2 || j == 3 || j == 7 || j == 5) {
      mark[j]++;
    } else if (j == 4) {
      mark[2] += 2;
      mark[3]++;
    } else if (j == 6) {
      mark[3]++;
      mark[5]++;
    } else if (j == 8) {
      mark[2] += 3;
      mark[7]++;
    } else if (j == 9) {
      mark[3] += 2;
      mark[2]++;
      mark[7]++;
    }
  }
  for (long long int i = 9; i >= 2; i--) {
    for (long long int j = 0; j < mark[i]; j++) {
      cout << i;
    }
  }
  cout << "\n";
}
int main() {
  long long int t = 1;
  while (t--) {
    solve();
  }
  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 mxn = 2e5 + 5;
void solve() {
  long long int n, k, m;
  long long int w;
  cin >> n;
  string s;
  cin >> s;
  long long int mark[12] = {0};
  for (long long int i = 0; i < n; i++) {
    int j = s[i] - '0';
    if (j == 1 || j == 0) continue;
    if (j == 2 || j == 3 || j == 7 || j == 5) {
      mark[j]++;
    } else if (j == 4) {
      mark[2] += 2;
      mark[3]++;
    } else if (j == 6) {
      mark[3]++;
      mark[5]++;
    } else if (j == 8) {
      mark[2] += 3;
      mark[7]++;
    } else if (j == 9) {
      mark[3] += 2;
      mark[2]++;
      mark[7]++;
    }
  }
  for (long long int i = 9; i >= 2; i--) {
    for (long long int j = 0; j < mark[i]; j++) {
      cout << i;
    }
  }
  cout << "\n";
}
int main() {
  long long int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```