#include <bits/stdc++.h>
using namespace std;
string str;
int N, i, k, j;
int cnt[10];
int main() {
  int N;
  cin >> N >> str;
  for (i = 0; i < N; ++i) {
    k = str[i] - '0';
    switch (k) {
      case 0:
        ++cnt[0];
        break;
      case 1:
        ++cnt[0];
        break;
      case 2:
        ++cnt[2];
        break;
      case 3:
        ++cnt[3];
        break;
      case 4:
        cnt[2] += 2, ++cnt[3];
        break;
      case 5:
        ++cnt[5];
        break;
      case 6:
        ++cnt[5], ++cnt[3];
        break;
      case 7:
        ++cnt[7];
        break;
      case 8:
        cnt[2] += 3, ++cnt[7];
        break;
      case 9:
        cnt[3] += 2, ++cnt[2], ++cnt[7];
    }
  }
  for (i = 9; i >= 1; --i)
    for (j = 1; j <= cnt[i]; ++j) cout << i;
  cout << endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
string str;
int N, i, k, j;
int cnt[10];
int main() {
  int N;
  cin >> N >> str;
  for (i = 0; i < N; ++i) {
    k = str[i] - '0';
    switch (k) {
      case 0:
        ++cnt[0];
        break;
      case 1:
        ++cnt[0];
        break;
      case 2:
        ++cnt[2];
        break;
      case 3:
        ++cnt[3];
        break;
      case 4:
        cnt[2] += 2, ++cnt[3];
        break;
      case 5:
        ++cnt[5];
        break;
      case 6:
        ++cnt[5], ++cnt[3];
        break;
      case 7:
        ++cnt[7];
        break;
      case 8:
        cnt[2] += 3, ++cnt[7];
        break;
      case 9:
        cnt[3] += 2, ++cnt[2], ++cnt[7];
    }
  }
  for (i = 9; i >= 1; --i)
    for (j = 1; j <= cnt[i]; ++j) cout << i;
  cout << endl;
  return 0;
}

```