#include <bits/stdc++.h>
using namespace std;
unsigned long long int power(unsigned long long int x,
                             unsigned long long int n) {
  if (n == 0) return 1;
  if (n % 2 == 0) {
    unsigned long long int y = power(x, n / 2);
    return (y * y);
  } else
    return ((x * pow(x, n - 1)));
}
int main() {
  int n, i, j;
  cin >> n;
  int f[10] = {0};
  long long int a;
  cin >> a;
  while (a) {
    f[a % 10]++;
    a /= 10;
  }
  if (f[4]) {
    f[2] += f[4] * 2;
    f[3] += f[4];
    f[4] = 0;
  }
  if (f[6]) {
    f[3] += f[6];
    f[5] += f[6];
    f[6] = 0;
  }
  if (f[8]) {
    f[2] += 3 * f[8];
    f[7] += f[8];
    f[8] = 0;
  }
  if (f[9]) {
    f[3] += 2 * f[9];
    f[2] += f[9];
    f[7] += f[9];
    f[9] = 0;
  }
  for (i = 7; i >= 2; i--) {
    if (f[i]) {
      for (j = 0; j < f[i]; j++) cout << i;
    }
  }
  cout << endl;
  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;
unsigned long long int power(unsigned long long int x,
                             unsigned long long int n) {
  if (n == 0) return 1;
  if (n % 2 == 0) {
    unsigned long long int y = power(x, n / 2);
    return (y * y);
  } else
    return ((x * pow(x, n - 1)));
}
int main() {
  int n, i, j;
  cin >> n;
  int f[10] = {0};
  long long int a;
  cin >> a;
  while (a) {
    f[a % 10]++;
    a /= 10;
  }
  if (f[4]) {
    f[2] += f[4] * 2;
    f[3] += f[4];
    f[4] = 0;
  }
  if (f[6]) {
    f[3] += f[6];
    f[5] += f[6];
    f[6] = 0;
  }
  if (f[8]) {
    f[2] += 3 * f[8];
    f[7] += f[8];
    f[8] = 0;
  }
  if (f[9]) {
    f[3] += 2 * f[9];
    f[2] += f[9];
    f[7] += f[9];
    f[9] = 0;
  }
  for (i = 7; i >= 2; i--) {
    if (f[i]) {
      for (j = 0; j < f[i]; j++) cout << i;
    }
  }
  cout << endl;
  return 0;
}

```