#include <bits/stdc++.h>
using namespace std;
const long long INF = 1E18;
const long double EPS = 1E-10;
const int N = 1e5 + 1;
const long long MOD = 1E9 + 7;
int n;
vector<char> v;
int main() {
  cin >> n;
  char a;
  for (int i = 1; i <= n; i++) {
    cin >> a;
    if (a == '4') {
      v.push_back('3');
      v.push_back('2');
      v.push_back('2');
    } else if (a == '6') {
      v.push_back('5');
      v.push_back('3');
    } else if (a == '8') {
      v.push_back('2');
      v.push_back('2');
      v.push_back('2');
      v.push_back('7');
    } else if (a == '9') {
      v.push_back('3');
      v.push_back('3');
      v.push_back('2');
      v.push_back('7');
    } else if (a != '0' && a != '1')
      v.push_back(a);
  }
  sort(v.begin(), v.end());
  for (int i = v.size() - 1; i >= 0; i--) cout << v[i];
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 long long INF = 1E18;
const long double EPS = 1E-10;
const int N = 1e5 + 1;
const long long MOD = 1E9 + 7;
int n;
vector<char> v;
int main() {
  cin >> n;
  char a;
  for (int i = 1; i <= n; i++) {
    cin >> a;
    if (a == '4') {
      v.push_back('3');
      v.push_back('2');
      v.push_back('2');
    } else if (a == '6') {
      v.push_back('5');
      v.push_back('3');
    } else if (a == '8') {
      v.push_back('2');
      v.push_back('2');
      v.push_back('2');
      v.push_back('7');
    } else if (a == '9') {
      v.push_back('3');
      v.push_back('3');
      v.push_back('2');
      v.push_back('7');
    } else if (a != '0' && a != '1')
      v.push_back(a);
  }
  sort(v.begin(), v.end());
  for (int i = v.size() - 1; i >= 0; i--) cout << v[i];
  return 0;
}

```