#include <bits/stdc++.h>
using namespace std;
int n;
char s[100];
int c[100];
int a[100];
int b[100];
int sol() {
  if (scanf("%d", &n) == EOF) return 0;
  scanf("%s", s);
  for (int i = 0; i < 10; i++) c[i] = a[i] = 0;
  for (int i = 0; i < n; i++) {
    int x = s[i] - '0';
    for (int j = 2; j <= x; j++) {
      if (j == 9) a[3] += 2;
      if (j == 8) a[2] += 3;
      if (j == 7) a[7] += 1;
      if (j == 6) {
        a[3] += 1;
        a[2] += 1;
      }
      if (j == 5) a[5] += 1;
      if (j == 4) a[2] += 2;
      if (j == 3) a[3] += 1;
      if (j == 2) a[2] += 1;
    }
  }
  for (; a[7] && a[5] && a[3] >= 2 && a[2] >= 4;) {
    printf("7");
    a[7]--;
    a[5]--;
    a[3] -= 2;
    a[2] -= 4;
  }
  for (; a[5] && a[3] && a[2] >= 3;) {
    printf("5");
    a[5]--;
    a[3]--;
    a[2] -= 3;
  }
  for (; a[3] && a[2];) {
    printf("3");
    a[3]--;
    a[2]--;
  }
  for (; a[2];) {
    printf("2");
    a[2]--;
  }
  puts("");
  return 1;
}
int main() {
  while (sol())
    ;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
char s[100];
int c[100];
int a[100];
int b[100];
int sol() {
  if (scanf("%d", &n) == EOF) return 0;
  scanf("%s", s);
  for (int i = 0; i < 10; i++) c[i] = a[i] = 0;
  for (int i = 0; i < n; i++) {
    int x = s[i] - '0';
    for (int j = 2; j <= x; j++) {
      if (j == 9) a[3] += 2;
      if (j == 8) a[2] += 3;
      if (j == 7) a[7] += 1;
      if (j == 6) {
        a[3] += 1;
        a[2] += 1;
      }
      if (j == 5) a[5] += 1;
      if (j == 4) a[2] += 2;
      if (j == 3) a[3] += 1;
      if (j == 2) a[2] += 1;
    }
  }
  for (; a[7] && a[5] && a[3] >= 2 && a[2] >= 4;) {
    printf("7");
    a[7]--;
    a[5]--;
    a[3] -= 2;
    a[2] -= 4;
  }
  for (; a[5] && a[3] && a[2] >= 3;) {
    printf("5");
    a[5]--;
    a[3]--;
    a[2] -= 3;
  }
  for (; a[3] && a[2];) {
    printf("3");
    a[3]--;
    a[2]--;
  }
  for (; a[2];) {
    printf("2");
    a[2]--;
  }
  puts("");
  return 1;
}
int main() {
  while (sol())
    ;
  return 0;
}

```