#include <bits/stdc++.h>
using namespace std;
const int maxint = -1u >> 1;
int n;
int sum[10];
char s[110];
int main() {
  scanf("%d", &n);
  scanf("%s", s);
  for (int i = 0; s[i]; i++) {
    int x = s[i] - '0';
    if (x == 2 || x == 3 || x == 5) sum[x]++;
    if (x == 4) sum[2] += 2, sum[3]++;
    if (x == 6) sum[5]++, sum[3]++;
    if (x == 7) sum[7]++;
    if (x == 8) sum[7]++, sum[2] += 3;
    if (x == 9) sum[7]++, sum[3] += 2, sum[2]++;
  }
  for (int i = 9; i >= 2; i--) {
    for (int j = 0; j < sum[i]; j++) {
      cout << i;
    }
  }
  cout << endl;
  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;
const int maxint = -1u >> 1;
int n;
int sum[10];
char s[110];
int main() {
  scanf("%d", &n);
  scanf("%s", s);
  for (int i = 0; s[i]; i++) {
    int x = s[i] - '0';
    if (x == 2 || x == 3 || x == 5) sum[x]++;
    if (x == 4) sum[2] += 2, sum[3]++;
    if (x == 6) sum[5]++, sum[3]++;
    if (x == 7) sum[7]++;
    if (x == 8) sum[7]++, sum[2] += 3;
    if (x == 9) sum[7]++, sum[3] += 2, sum[2]++;
  }
  for (int i = 9; i >= 2; i--) {
    for (int j = 0; j < sum[i]; j++) {
      cout << i;
    }
  }
  cout << endl;
  return 0;
}

```