#include <bits/stdc++.h>
using namespace std;
void start() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
int main() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  int a[10];
  for (int i = 0; i < 10; i++) a[i] = 0;
  for (int i = 0; i < n; i++)
    for (int j = 2; j <= (s[i] - '0'); j++) a[j]++;
  for (int j = 0; j < a[7]; j++)
    for (int h = 6; h > 1; h--) a[h]--;
  for (int j = 0; j < a[5]; j++)
    for (int h = 4; h > 1; h--) a[h]--;
  a[3] += 2 * a[9] + a[6];
  a[9] = 0;
  a[2] += 3 * a[8] + 2 * a[4] + a[6];
  a[8] = 0;
  a[6] = 0;
  a[4] = 0;
  a[2] -= a[3];
  for (int i = 9; i > 1; i--)
    for (int j = 0; j < a[i]; j++) cout << i;
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
void start() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
int main() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  int a[10];
  for (int i = 0; i < 10; i++) a[i] = 0;
  for (int i = 0; i < n; i++)
    for (int j = 2; j <= (s[i] - '0'); j++) a[j]++;
  for (int j = 0; j < a[7]; j++)
    for (int h = 6; h > 1; h--) a[h]--;
  for (int j = 0; j < a[5]; j++)
    for (int h = 4; h > 1; h--) a[h]--;
  a[3] += 2 * a[9] + a[6];
  a[9] = 0;
  a[2] += 3 * a[8] + 2 * a[4] + a[6];
  a[8] = 0;
  a[6] = 0;
  a[4] = 0;
  a[2] -= a[3];
  for (int i = 9; i > 1; i--)
    for (int j = 0; j < a[i]; j++) cout << i;
  return 0;
}

```