#include <bits/stdc++.h>
using namespace std;
vector<int> aux[15];
int p[15];
void add(int num, int val) {
  for (int i = 2; i < num + 1; i++) {
    for (int j = 0; j < aux[i].size(); j++) {
      p[aux[i][j]] += val;
    }
  }
}
int main() {
  for (int i = 2; i < 10; i++) {
    int num = i;
    for (int j = 2; j < 10; j++) {
      while (num % j == 0) {
        num /= j;
        aux[i].push_back(j);
      }
    }
  }
  memset(p, 0, sizeof p);
  int n;
  cin >> n;
  string a;
  cin >> a;
  for (int i = 0; i < n; i++) add(a[i] - '0', 1);
  for (int i = 7; i >= 2; i--) {
    while (p[i] > 0) {
      cout << i;
      add(i, -1);
    }
  }
  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;
vector<int> aux[15];
int p[15];
void add(int num, int val) {
  for (int i = 2; i < num + 1; i++) {
    for (int j = 0; j < aux[i].size(); j++) {
      p[aux[i][j]] += val;
    }
  }
}
int main() {
  for (int i = 2; i < 10; i++) {
    int num = i;
    for (int j = 2; j < 10; j++) {
      while (num % j == 0) {
        num /= j;
        aux[i].push_back(j);
      }
    }
  }
  memset(p, 0, sizeof p);
  int n;
  cin >> n;
  string a;
  cin >> a;
  for (int i = 0; i < n; i++) add(a[i] - '0', 1);
  for (int i = 7; i >= 2; i--) {
    while (p[i] > 0) {
      cout << i;
      add(i, -1);
    }
  }
  cout << endl;
  return 0;
}

```