#include <bits/stdc++.h>
using namespace std;
int jc[15];
char s[20];
int ss[50];
int main() {
  jc[0] = 1;
  for (int i = 1; i < 10; i++) {
    jc[i] = jc[i - 1] * i;
  }
  int n;
  scanf("%d", &n);
  scanf("%s", s);
  long long ans = 1;
  int k = 0;
  for (int i = 0; i < n; i++) {
    ans = jc[s[i] - '0'];
    while (ans % jc[7] == 0) {
      ss[k++] = 7;
      ans /= jc[7];
    }
    while (ans % jc[5] == 0) {
      ss[k++] = 5;
      ans /= jc[5];
    }
    while (ans % jc[3] == 0) {
      ss[k++] = 3;
      ans /= jc[3];
    }
    while (ans % jc[2] == 0) {
      ss[k++] = 2;
      ans /= jc[2];
    }
  }
  sort(ss, ss + k);
  for (int i = k - 1; i >= 0; i--) printf("%d", ss[i]);
  printf("\n");
}

### Prompt

Your task is to create 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 jc[15];
char s[20];
int ss[50];
int main() {
  jc[0] = 1;
  for (int i = 1; i < 10; i++) {
    jc[i] = jc[i - 1] * i;
  }
  int n;
  scanf("%d", &n);
  scanf("%s", s);
  long long ans = 1;
  int k = 0;
  for (int i = 0; i < n; i++) {
    ans = jc[s[i] - '0'];
    while (ans % jc[7] == 0) {
      ss[k++] = 7;
      ans /= jc[7];
    }
    while (ans % jc[5] == 0) {
      ss[k++] = 5;
      ans /= jc[5];
    }
    while (ans % jc[3] == 0) {
      ss[k++] = 3;
      ans /= jc[3];
    }
    while (ans % jc[2] == 0) {
      ss[k++] = 2;
      ans /= jc[2];
    }
  }
  sort(ss, ss + k);
  for (int i = k - 1; i >= 0; i--) printf("%d", ss[i]);
  printf("\n");
}

```