#include <bits/stdc++.h>
using namespace std;
string s;
int n;
int A[10], B[10];
int main() {
  A[0] = 2;
  A[1] = 3;
  A[2] = 5;
  A[3] = 7;
  cin >> n;
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    int a = s[i] - '0';
    for (int j = 2; j <= a; j++) {
      int x = j;
      for (int k = 0; k < 4; k++) {
        while (x % A[k] == 0) B[k]++, x /= A[k];
      }
    }
  }
  vector<int> ans;
  for (int i = 3; i >= 0; i--) {
    int num = B[i];
    for (int t = 0; t < num; t++) ans.push_back(A[i]);
    for (int j = 0; j < 4; j++) {
      int a = A[j];
      int cnt = 0;
      for (int t = 1; t <= A[i]; t++) {
        int x = t;
        while (x % a == 0) x /= a, cnt++;
      }
      B[j] -= cnt * num;
    }
  }
  sort((ans).begin(), (ans).end());
  reverse((ans).begin(), (ans).end());
  for (int i = 0; i < ans.size(); i++) cout << ans[i];
  cout << '\n';
}

### 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;
string s;
int n;
int A[10], B[10];
int main() {
  A[0] = 2;
  A[1] = 3;
  A[2] = 5;
  A[3] = 7;
  cin >> n;
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    int a = s[i] - '0';
    for (int j = 2; j <= a; j++) {
      int x = j;
      for (int k = 0; k < 4; k++) {
        while (x % A[k] == 0) B[k]++, x /= A[k];
      }
    }
  }
  vector<int> ans;
  for (int i = 3; i >= 0; i--) {
    int num = B[i];
    for (int t = 0; t < num; t++) ans.push_back(A[i]);
    for (int j = 0; j < 4; j++) {
      int a = A[j];
      int cnt = 0;
      for (int t = 1; t <= A[i]; t++) {
        int x = t;
        while (x % a == 0) x /= a, cnt++;
      }
      B[j] -= cnt * num;
    }
  }
  sort((ans).begin(), (ans).end());
  reverse((ans).begin(), (ans).end());
  for (int i = 0; i < ans.size(); i++) cout << ans[i];
  cout << '\n';
}

```