#include <bits/stdc++.h>
using namespace std;
int n;
int dd[20];
char p;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p;
    int u = p - '0';
    if (u == 1) continue;
    if (u == 4) {
      dd[2] += 2;
      dd[3]++;
      continue;
    }
    if (u == 6) {
      dd[3]++;
      dd[5]++;
      continue;
    }
    if (u == 8) {
      dd[7]++;
      dd[2] += 3;
      continue;
    }
    if (u == 9) {
      dd[3] += 2;
      dd[7]++;
      dd[2]++;
      continue;
    }
    dd[u]++;
  }
  for (int i = 9; i >= 1; i--)
    for (int j = 1; j <= dd[i]; j++) cout << i;
}

### 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 n;
int dd[20];
char p;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p;
    int u = p - '0';
    if (u == 1) continue;
    if (u == 4) {
      dd[2] += 2;
      dd[3]++;
      continue;
    }
    if (u == 6) {
      dd[3]++;
      dd[5]++;
      continue;
    }
    if (u == 8) {
      dd[7]++;
      dd[2] += 3;
      continue;
    }
    if (u == 9) {
      dd[3] += 2;
      dd[7]++;
      dd[2]++;
      continue;
    }
    dd[u]++;
  }
  for (int i = 9; i >= 1; i--)
    for (int j = 1; j <= dd[i]; j++) cout << i;
}

```