#include <bits/stdc++.h>
using namespace std;
const int maxN = 1000 * 1000 + 5;
const int mod = 1000 * 1000 * 1000 + 7;
const int maxL = 20;
vector<int> v[10];
int cnt[maxN];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int a[10][10] = {
      {0},    {0},       {1, 2}, {1, 3},          {3, 2, 2, 3},
      {1, 5}, {2, 5, 3}, {1, 7}, {4, 7, 2, 2, 2}, {4, 7, 3, 3, 2},
  };
  int n;
  cin >> n;
  string s;
  cin >> s;
  for (int i = 0; i < (((int)s.size())); i++) {
    cnt[s[i] - '0']++;
  }
  vector<int> ans;
  for (int i = 2; i <= 9; i++) {
    for (int j = 0; j < cnt[i]; j++) {
      int te = a[i][0];
      for (int k = 1; k <= te; k++) ans.push_back(a[i][k]);
    }
  }
  sort(ans.begin(), ans.end(), greater<int>());
  for (auto u : ans) cout << u;
  cout << endl;
  return 0;
}

### Prompt

Please provide a CPP coded solution 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 int maxN = 1000 * 1000 + 5;
const int mod = 1000 * 1000 * 1000 + 7;
const int maxL = 20;
vector<int> v[10];
int cnt[maxN];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int a[10][10] = {
      {0},    {0},       {1, 2}, {1, 3},          {3, 2, 2, 3},
      {1, 5}, {2, 5, 3}, {1, 7}, {4, 7, 2, 2, 2}, {4, 7, 3, 3, 2},
  };
  int n;
  cin >> n;
  string s;
  cin >> s;
  for (int i = 0; i < (((int)s.size())); i++) {
    cnt[s[i] - '0']++;
  }
  vector<int> ans;
  for (int i = 2; i <= 9; i++) {
    for (int j = 0; j < cnt[i]; j++) {
      int te = a[i][0];
      for (int k = 1; k <= te; k++) ans.push_back(a[i][k]);
    }
  }
  sort(ans.begin(), ans.end(), greater<int>());
  for (auto u : ans) cout << u;
  cout << endl;
  return 0;
}

```