#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int N;
  cin >> N;
  int a[10], i, d;
  for (i = 0; i < 10; ++i) {
    a[i] = 0;
  }
  cin >> N;
  while (N) {
    d = N % 10;
    if (d == 4) {
      ++a[3];
      a[2] += 2;
    } else if (d == 6) {
      ++a[5];
      ++a[3];
    } else if (d == 8) {
      ++a[7];
      a[2] += 3;
    } else if (d == 9) {
      ++a[7];
      a[2] += 1;
      a[3] += 2;
    } else
      ++a[d];
    N = N / 10;
  }
  for (i = 7; i > 1; --i) {
    while (a[i]) {
      cout << i;
      --a[i];
    }
  }
}

### 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 main() {
  long long int N;
  cin >> N;
  int a[10], i, d;
  for (i = 0; i < 10; ++i) {
    a[i] = 0;
  }
  cin >> N;
  while (N) {
    d = N % 10;
    if (d == 4) {
      ++a[3];
      a[2] += 2;
    } else if (d == 6) {
      ++a[5];
      ++a[3];
    } else if (d == 8) {
      ++a[7];
      a[2] += 3;
    } else if (d == 9) {
      ++a[7];
      a[2] += 1;
      a[3] += 2;
    } else
      ++a[d];
    N = N / 10;
  }
  for (i = 7; i > 1; --i) {
    while (a[i]) {
      cout << i;
      --a[i];
    }
  }
}

```