#include <bits/stdc++.h>
using namespace std;
int cnt[15];
int extra2, extra3;
int n;
string s;
int main() {
  cin >> n >> s;
  for (int i = 0; i < n; i++) {
    if (s[i] == '0' || s[i] == '1') continue;
    if (s[i] == '4') {
      cnt[3]++;
      extra2 += 2;
    } else if (s[i] == '6') {
      cnt[5]++;
      extra3++;
    } else if (s[i] == '8') {
      cnt[7]++;
      extra2 += 3;
    } else if (s[i] == '9') {
      cnt[7]++;
      extra3 += 2;
      extra2 += 1;
    } else {
      cnt[s[i] - '0']++;
    }
  }
  cnt[2] += extra2, cnt[3] += extra3;
  for (int i = 9; i >= 2; i--) {
    while (cnt[i]) {
      cout << i;
      cnt[i]--;
    }
  }
  puts("");
  return 0;
}

### Prompt

Please create a solution in Cpp 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 cnt[15];
int extra2, extra3;
int n;
string s;
int main() {
  cin >> n >> s;
  for (int i = 0; i < n; i++) {
    if (s[i] == '0' || s[i] == '1') continue;
    if (s[i] == '4') {
      cnt[3]++;
      extra2 += 2;
    } else if (s[i] == '6') {
      cnt[5]++;
      extra3++;
    } else if (s[i] == '8') {
      cnt[7]++;
      extra2 += 3;
    } else if (s[i] == '9') {
      cnt[7]++;
      extra3 += 2;
      extra2 += 1;
    } else {
      cnt[s[i] - '0']++;
    }
  }
  cnt[2] += extra2, cnt[3] += extra3;
  for (int i = 9; i >= 2; i--) {
    while (cnt[i]) {
      cout << i;
      cnt[i]--;
    }
  }
  puts("");
  return 0;
}

```