#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  int f[4] = {2, 3, 5, 7};
  map<int, int> cnt;
  vector<int> ans;
  for (int i = 0; i < n; ++i) {
    int d = s[i] - '0';
    for (int j = 2; j <= d; ++j) {
      int a = j;
      for (int k = 0; k < 4; ++k)
        while (a % f[k] == 0 and a > 1) ++cnt[f[k]], a /= f[k];
    }
  }
  while (cnt[7]) {
    ans.push_back(7);
    --cnt[7], --cnt[5], cnt[3] -= 2, cnt[2] -= 4;
  }
  while (cnt[5]) {
    ans.push_back(5);
    --cnt[5], --cnt[3], cnt[2] -= 3;
  }
  while (cnt[3]) {
    ans.push_back(3);
    --cnt[3], --cnt[2];
  }
  for (int i = 0; i < cnt[2]; ++i) ans.push_back(2);
  sort(ans.rbegin(), ans.rend());
  copy(ans.begin(), ans.end(), ostream_iterator<int>(cout, ""));
  return 0;
}

### Prompt

Please formulate 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;
int main() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  int f[4] = {2, 3, 5, 7};
  map<int, int> cnt;
  vector<int> ans;
  for (int i = 0; i < n; ++i) {
    int d = s[i] - '0';
    for (int j = 2; j <= d; ++j) {
      int a = j;
      for (int k = 0; k < 4; ++k)
        while (a % f[k] == 0 and a > 1) ++cnt[f[k]], a /= f[k];
    }
  }
  while (cnt[7]) {
    ans.push_back(7);
    --cnt[7], --cnt[5], cnt[3] -= 2, cnt[2] -= 4;
  }
  while (cnt[5]) {
    ans.push_back(5);
    --cnt[5], --cnt[3], cnt[2] -= 3;
  }
  while (cnt[3]) {
    ans.push_back(3);
    --cnt[3], --cnt[2];
  }
  for (int i = 0; i < cnt[2]; ++i) ans.push_back(2);
  sort(ans.rbegin(), ans.rend());
  copy(ans.begin(), ans.end(), ostream_iterator<int>(cout, ""));
  return 0;
}

```