#include <bits/stdc++.h>
const int p[10][4] = {{0, 0, 0, 0}, {0, 0, 0, 0}, {1, 0, 0, 0}, {1, 1, 0, 0},
                      {3, 1, 0, 0}, {3, 1, 1, 0}, {4, 2, 1, 0}, {4, 2, 1, 1},
                      {7, 2, 1, 1}, {7, 4, 1, 1}};
int main() {
  int n;
  scanf("%d", &n);
  char s[n + 1];
  scanf("%s", s);
  int sp[4];
  memset(sp, 0, sizeof(sp));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 4; ++j) {
      sp[j] += p[s[i] - '0'][j];
    }
  }
  while (sp[3]) {
    putchar('7');
    sp[0] -= 4;
    sp[1] -= 2;
    sp[2] -= 1;
    sp[3] -= 1;
  }
  while (sp[2]) {
    putchar('5');
    sp[0] -= 3;
    sp[1] -= 1;
    sp[2] -= 1;
  }
  while (sp[1]) {
    putchar('3');
    sp[0] -= 1;
    sp[1] -= 1;
  }
  while (sp[0]) {
    putchar('2');
    sp[0] -= 1;
  }
  printf("\n");
  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>
const int p[10][4] = {{0, 0, 0, 0}, {0, 0, 0, 0}, {1, 0, 0, 0}, {1, 1, 0, 0},
                      {3, 1, 0, 0}, {3, 1, 1, 0}, {4, 2, 1, 0}, {4, 2, 1, 1},
                      {7, 2, 1, 1}, {7, 4, 1, 1}};
int main() {
  int n;
  scanf("%d", &n);
  char s[n + 1];
  scanf("%s", s);
  int sp[4];
  memset(sp, 0, sizeof(sp));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 4; ++j) {
      sp[j] += p[s[i] - '0'][j];
    }
  }
  while (sp[3]) {
    putchar('7');
    sp[0] -= 4;
    sp[1] -= 2;
    sp[2] -= 1;
    sp[3] -= 1;
  }
  while (sp[2]) {
    putchar('5');
    sp[0] -= 3;
    sp[1] -= 1;
    sp[2] -= 1;
  }
  while (sp[1]) {
    putchar('3');
    sp[0] -= 1;
    sp[1] -= 1;
  }
  while (sp[0]) {
    putchar('2');
    sp[0] -= 1;
  }
  printf("\n");
  return 0;
}

```