#include <bits/stdc++.h>
using namespace std;
int fact[10] = {1};
int f[10];
long long num;
void fac(int x, int a) {
  int i = 2;
  while (x != 1) {
    while (x % i == 0) {
      f[i] += a;
      x /= i;
    }
    i++;
  }
}
void init() {
  for (int i = 1; i < 10; i++) fact[i] = fact[i - 1] * i;
}
bool check() {
  for (auto i : f)
    if (i) return false;
  return true;
}
int main() {
  init();
  int d;
  cin >> d;
  cin >> num;
  while (num) {
    fac(fact[num % 10], 1);
    num /= 10;
  }
  int ind = 9;
  vector<int> ans;
  while (!check()) {
    while (f[ind]) {
      ans.push_back(ind);
      fac(fact[ind], -1);
    }
    ind--;
  }
  for (auto i : ans) cout << i;
  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;
int fact[10] = {1};
int f[10];
long long num;
void fac(int x, int a) {
  int i = 2;
  while (x != 1) {
    while (x % i == 0) {
      f[i] += a;
      x /= i;
    }
    i++;
  }
}
void init() {
  for (int i = 1; i < 10; i++) fact[i] = fact[i - 1] * i;
}
bool check() {
  for (auto i : f)
    if (i) return false;
  return true;
}
int main() {
  init();
  int d;
  cin >> d;
  cin >> num;
  while (num) {
    fac(fact[num % 10], 1);
    num /= 10;
  }
  int ind = 9;
  vector<int> ans;
  while (!check()) {
    while (f[ind]) {
      ans.push_back(ind);
      fac(fact[ind], -1);
    }
    ind--;
  }
  for (auto i : ans) cout << i;
  cout << endl;
  return 0;
}

```