#include <bits/stdc++.h>
using namespace std;
int a[10][4] = {{0, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 1}, {0, 0, 1, 1},
                {0, 0, 1, 3}, {0, 1, 1, 3}, {0, 1, 2, 4}, {1, 1, 2, 4},
                {1, 1, 2, 7}, {1, 1, 4, 7}},
    x[4];
int main() {
  string s;
  int n;
  cin >> n >> s;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 4; j++) x[j] += a[s[i] - '0'][j];
  for (int i = 0; i < x[0]; i++)
    cout << '7', x[1] -= a[6][1], x[2] -= a[6][2], x[3] -= a[6][3];
  for (int i = 0; i < x[1]; i++) cout << '5', x[2] -= a[4][2], x[3] -= a[4][3];
  for (int i = 0; i < x[2]; i++) cout << '3', x[3] -= a[2][3];
  for (int i = 0; i < x[3]; i++) cout << '2';
}

### Prompt

In cpp, your task is to solve 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 a[10][4] = {{0, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 1}, {0, 0, 1, 1},
                {0, 0, 1, 3}, {0, 1, 1, 3}, {0, 1, 2, 4}, {1, 1, 2, 4},
                {1, 1, 2, 7}, {1, 1, 4, 7}},
    x[4];
int main() {
  string s;
  int n;
  cin >> n >> s;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 4; j++) x[j] += a[s[i] - '0'][j];
  for (int i = 0; i < x[0]; i++)
    cout << '7', x[1] -= a[6][1], x[2] -= a[6][2], x[3] -= a[6][3];
  for (int i = 0; i < x[1]; i++) cout << '5', x[2] -= a[4][2], x[3] -= a[4][3];
  for (int i = 0; i < x[2]; i++) cout << '3', x[3] -= a[2][3];
  for (int i = 0; i < x[3]; i++) cout << '2';
}

```