#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, v[100] = {0};
  long long a, c, m[100][100] = {{},  {},     {2}, {3},       {2, 2},
                                 {5}, {2, 3}, {7}, {2, 2, 2}, {3, 3}};
  long long b = 1;
  cin >> n >> a;
  for (int i = 0; i < n; i++) {
    c = a % 10;
    for (int k = 2; k <= c; k++) {
      for (int l = 0; l < 3; l++) {
        if (m[k][l] != 0) {
          v[m[k][l]]++;
        }
      }
    }
    a /= 10;
  }
  for (int i = 9; i >= 2; i--) {
    for (int j = 0; j < v[i]; j++) {
      cout << i;
      for (int k = i - 1; k > 0; k--) {
        for (int l = 0; l < 3; l++) {
          v[m[k][l]]--;
        }
      }
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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 main() {
  int n, v[100] = {0};
  long long a, c, m[100][100] = {{},  {},     {2}, {3},       {2, 2},
                                 {5}, {2, 3}, {7}, {2, 2, 2}, {3, 3}};
  long long b = 1;
  cin >> n >> a;
  for (int i = 0; i < n; i++) {
    c = a % 10;
    for (int k = 2; k <= c; k++) {
      for (int l = 0; l < 3; l++) {
        if (m[k][l] != 0) {
          v[m[k][l]]++;
        }
      }
    }
    a /= 10;
  }
  for (int i = 9; i >= 2; i--) {
    for (int j = 0; j < v[i]; j++) {
      cout << i;
      for (int k = i - 1; k > 0; k--) {
        for (int l = 0; l < 3; l++) {
          v[m[k][l]]--;
        }
      }
    }
  }
  return 0;
}

```