#include <bits/stdc++.h>
int main() {
  int n;
  char a[20];
  scanf(" %d", &n);
  scanf(" %s", a);
  char num[10];
  for (int i = 2; i < 10; i++) {
    int c = 0;
    for (int j = 0; a[j]; j++)
      if (a[j] >= i + '0') c++;
    num[i] = c;
  }
  if (num[7] > 0) {
    for (int i = 2; i < 7; i++) {
      num[i] -= num[7];
    }
    while (num[7]--) printf("7");
  }
  if (num[5] > 0) {
    for (int i = 2; i < 5; i++) {
      num[i] -= num[5];
    }
    while (num[5]--) printf("5");
  }
  num[3] += num[6];
  num[3] += num[9] * 2;
  num[2] += num[6];
  num[2] += num[4] * 2;
  num[2] += num[8] * 3;
  num[2] -= num[3];
  while (num[3]--) printf("3");
  while (num[2]--) printf("2");
  printf("\n");
  return 0;
}

### Prompt

Generate 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 main() {
  int n;
  char a[20];
  scanf(" %d", &n);
  scanf(" %s", a);
  char num[10];
  for (int i = 2; i < 10; i++) {
    int c = 0;
    for (int j = 0; a[j]; j++)
      if (a[j] >= i + '0') c++;
    num[i] = c;
  }
  if (num[7] > 0) {
    for (int i = 2; i < 7; i++) {
      num[i] -= num[7];
    }
    while (num[7]--) printf("7");
  }
  if (num[5] > 0) {
    for (int i = 2; i < 5; i++) {
      num[i] -= num[5];
    }
    while (num[5]--) printf("5");
  }
  num[3] += num[6];
  num[3] += num[9] * 2;
  num[2] += num[6];
  num[2] += num[4] * 2;
  num[2] += num[8] * 3;
  num[2] -= num[3];
  while (num[3]--) printf("3");
  while (num[2]--) printf("2");
  printf("\n");
  return 0;
}

```