#include <bits/stdc++.h>
int count[10];
int main() {
  int n;
  char buf[100];
  char ans[500];
  int idx = 0;
  scanf("%d", &n);
  scanf("%s", buf);
  for (int j = 0; j < 10; j++) {
    count[j] = 0;
  }
  for (int i = 0; i < n; i++) {
    int j = buf[i] - '1' + 1;
    if (j == 9) {
      count[3] += 2;
      count[2]++;
      count[7]++;
    } else if (j == 8) {
      count[2] += 3;
      count[7]++;
    } else if (j == 7) {
      count[7]++;
    } else if (j == 6) {
      count[3]++;
      count[5]++;
    } else if (j == 5) {
      count[5]++;
    } else if (j == 4) {
      count[2] += 2;
      count[3]++;
    } else if (j <= 3) {
      count[j]++;
    }
  }
  for (int j = 7; j > 1; j--) {
    while (count[j] > 0) {
      ans[idx++] = j - 1 + '1';
      count[j]--;
    }
  }
  ans[idx] = '\0';
  printf("%s\n", ans);
  return 0;
}

### Prompt

Please formulate 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>
int count[10];
int main() {
  int n;
  char buf[100];
  char ans[500];
  int idx = 0;
  scanf("%d", &n);
  scanf("%s", buf);
  for (int j = 0; j < 10; j++) {
    count[j] = 0;
  }
  for (int i = 0; i < n; i++) {
    int j = buf[i] - '1' + 1;
    if (j == 9) {
      count[3] += 2;
      count[2]++;
      count[7]++;
    } else if (j == 8) {
      count[2] += 3;
      count[7]++;
    } else if (j == 7) {
      count[7]++;
    } else if (j == 6) {
      count[3]++;
      count[5]++;
    } else if (j == 5) {
      count[5]++;
    } else if (j == 4) {
      count[2] += 2;
      count[3]++;
    } else if (j <= 3) {
      count[j]++;
    }
  }
  for (int j = 7; j > 1; j--) {
    while (count[j] > 0) {
      ans[idx++] = j - 1 + '1';
      count[j]--;
    }
  }
  ans[idx] = '\0';
  printf("%s\n", ans);
  return 0;
}

```