#include <bits/stdc++.h>
using namespace std;
int n, t[20];
char ch;
bool check(int s) {
  for (int i = 2; i < s; i++)
    if (s % i == 0) return false;
  return true;
}
int main() {
  scanf("%d\n", &n);
  for (int i = 0; i < n; i++) {
    cin >> ch;
    int l = ch - '0';
    for (int i = 2; i <= l; i++) t[i]++;
  }
  int m = 0;
  for (int i = 9; i > 1; i--) {
    t[i] -= m;
    if (t[i] == 0) continue;
    if (check(i)) {
      for (int j = 0; j < t[i]; j++) printf("%d", i);
      m += t[i];
    } else {
      int gg = i;
      for (int j = 2; j < i; j++)
        while (gg % j == 0) {
          t[j] += t[i];
          gg /= j;
        }
    }
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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, t[20];
char ch;
bool check(int s) {
  for (int i = 2; i < s; i++)
    if (s % i == 0) return false;
  return true;
}
int main() {
  scanf("%d\n", &n);
  for (int i = 0; i < n; i++) {
    cin >> ch;
    int l = ch - '0';
    for (int i = 2; i <= l; i++) t[i]++;
  }
  int m = 0;
  for (int i = 9; i > 1; i--) {
    t[i] -= m;
    if (t[i] == 0) continue;
    if (check(i)) {
      for (int j = 0; j < t[i]; j++) printf("%d", i);
      m += t[i];
    } else {
      int gg = i;
      for (int j = 2; j < i; j++)
        while (gg % j == 0) {
          t[j] += t[i];
          gg /= j;
        }
    }
  }
  return 0;
}

```