#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:1000000000")
using namespace std;
const bool db = false;
int n;
string a;
int f[] = {2, 3, 5, 7};
int cnt[10];
string dp[10];
int main() {
  cin >> n;
  cin >> a;
  sort(a.begin(), a.end());
  while (*a.begin() <= 1) a.erase(a.begin());
  dp[2] = "2";
  dp[3] = "3";
  dp[4] = "223";
  dp[5] = "5";
  dp[6] = "53";
  dp[7] = "7";
  dp[8] = "2227";
  dp[9] = "2337";
  string res;
  for (int i = 0; i < a.size(); ++i) ++cnt[a[i] - '0'];
  for (int i = 2; i < 10; ++i) {
    for (int j = 0; j < cnt[i]; ++j) {
      res += dp[i];
    }
  }
  sort(res.begin(), res.end());
  reverse(res.begin(), res.end());
  cout << res << endl;
  getchar();
  getchar();
  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>
#pragma comment(linker, "/STACK:1000000000")
using namespace std;
const bool db = false;
int n;
string a;
int f[] = {2, 3, 5, 7};
int cnt[10];
string dp[10];
int main() {
  cin >> n;
  cin >> a;
  sort(a.begin(), a.end());
  while (*a.begin() <= 1) a.erase(a.begin());
  dp[2] = "2";
  dp[3] = "3";
  dp[4] = "223";
  dp[5] = "5";
  dp[6] = "53";
  dp[7] = "7";
  dp[8] = "2227";
  dp[9] = "2337";
  string res;
  for (int i = 0; i < a.size(); ++i) ++cnt[a[i] - '0'];
  for (int i = 2; i < 10; ++i) {
    for (int j = 0; j < cnt[i]; ++j) {
      res += dp[i];
    }
  }
  sort(res.begin(), res.end());
  reverse(res.begin(), res.end());
  cout << res << endl;
  getchar();
  getchar();
  return 0;
}

```