#include <bits/stdc++.h>
using namespace std;
int n;
char s[20];
int q[10];
int main() {
  cin >> n >> s;
  for (int i = 0; i < n; ++i) {
    s[i] -= '0';
    for (int j = 2; j <= s[i]; ++j) {
      int jj = j;
      while (jj % 2 == 0) ++q[2], jj /= 2;
      while (jj % 3 == 0) ++q[3], jj /= 3;
      while (jj % 5 == 0) ++q[5], jj /= 5;
      while (jj % 7 == 0) ++q[7], jj /= 7;
    }
  }
  int q7 = q[7];
  q[5] -= q7, q[3] -= 2 * q7, q[2] -= 4 * q7;
  int q5 = q[5];
  q[3] -= q5, q[2] -= 3 * q5;
  int q3 = q[3];
  q[2] -= q3;
  int q2 = q[2];
  for (int i = 0; i < q7; ++i) cout << 7;
  for (int i = 0; i < q5; ++i) cout << 5;
  for (int i = 0; i < q3; ++i) cout << 3;
  for (int i = 0; i < q2; ++i) cout << 2;
  cout << endl;
  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>
using namespace std;
int n;
char s[20];
int q[10];
int main() {
  cin >> n >> s;
  for (int i = 0; i < n; ++i) {
    s[i] -= '0';
    for (int j = 2; j <= s[i]; ++j) {
      int jj = j;
      while (jj % 2 == 0) ++q[2], jj /= 2;
      while (jj % 3 == 0) ++q[3], jj /= 3;
      while (jj % 5 == 0) ++q[5], jj /= 5;
      while (jj % 7 == 0) ++q[7], jj /= 7;
    }
  }
  int q7 = q[7];
  q[5] -= q7, q[3] -= 2 * q7, q[2] -= 4 * q7;
  int q5 = q[5];
  q[3] -= q5, q[2] -= 3 * q5;
  int q3 = q[3];
  q[2] -= q3;
  int q2 = q[2];
  for (int i = 0; i < q7; ++i) cout << 7;
  for (int i = 0; i < q5; ++i) cout << 5;
  for (int i = 0; i < q3; ++i) cout << 3;
  for (int i = 0; i < q2; ++i) cout << 2;
  cout << endl;
  return 0;
}

```