#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
using ll = long long;
using lld = long double;
using vll = int64_t;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  string second;
  cin >> second;
  unordered_map<char, vector<int>> hm;
  hm['2'] = {2};
  hm['3'] = {3};
  hm['4'] = {3, 2, 2};
  hm['5'] = {5};
  hm['6'] = {3, 5};
  hm['7'] = {7};
  hm['8'] = {2, 2, 2, 7};
  hm['9'] = {2, 3, 3, 7};
  vector<int> ans;
  for (auto i : second)
    for (int j : hm[i]) ans.push_back(j);
  sort(ans.begin(), ans.end(), greater<int>());
  for (int i : ans) cout << i;
  cout << '\n';
  return 0;
}

### Prompt

Generate 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>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
using ll = long long;
using lld = long double;
using vll = int64_t;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  string second;
  cin >> second;
  unordered_map<char, vector<int>> hm;
  hm['2'] = {2};
  hm['3'] = {3};
  hm['4'] = {3, 2, 2};
  hm['5'] = {5};
  hm['6'] = {3, 5};
  hm['7'] = {7};
  hm['8'] = {2, 2, 2, 7};
  hm['9'] = {2, 3, 3, 7};
  vector<int> ans;
  for (auto i : second)
    for (int j : hm[i]) ans.push_back(j);
  sort(ans.begin(), ans.end(), greater<int>());
  for (int i : ans) cout << i;
  cout << '\n';
  return 0;
}

```