#include <bits/stdc++.h>
unsigned long long int f(unsigned long long int n) {
  unsigned long long int p;
  p = 1;
  if (n == 0)
    return 1;
  else if (n > 0) {
    while (n > 0) {
      p = p * n;
      n--;
    }
    return p;
  }
}
int main() {
  unsigned long long int z, b, to;
  long long int n, i, c, j, d, r, h;
  c = 0, d = 0;
  unsigned long long int a[15];
  for (i = 1; i <= 15; i++) a[i] = 0;
  char s[15];
  z = 1;
  scanf("%lld", &n);
  scanf("%s", s);
  for (i = 0; i < n; i++) {
    if (s[i] == 48) c++;
    z = z * f(s[i] - 48);
  }
  j = n - 1;
  while (j >= 0) {
    if (f(s[j] - 48) % f(7) == 0) {
      a[7] = a[7] + 1;
      if (s[j] - 48 == 8) {
        a[2] = a[2] + 3;
      } else if (s[j] == 57) {
        a[3] = a[3] + 2;
        a[2] = a[2] + 3 - 2;
      }
    } else if (f(s[j] - 48) % f(5) == 0) {
      a[5] = a[5] + 1;
      if (s[j] == 54) {
        a[3] = a[3] + 1;
      }
    } else if (f(s[j] - 48) % f(3) == 0) {
      a[3] = a[3] + 1;
      if (s[j] == 52) a[2] = a[2] + 2;
    } else if (f(s[j] - 48) % f(2) == 0) {
      if (s[j] == 50) a[2] = a[2] + 1;
    }
    j--;
  }
  r = a[3] + a[2] + a[5] + a[7];
  while (a[7] > 0) {
    printf("7");
    a[7] = a[7] - 1;
  }
  while (a[5] > 0) {
    printf("5");
    a[5] = a[5] - 1;
  }
  while (a[3] > 0) {
    printf("3");
    a[3] = a[3] - 1;
  }
  while (a[2] > 0) {
    printf("2");
    a[2] = a[2] - 1;
  }
}

### 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>
unsigned long long int f(unsigned long long int n) {
  unsigned long long int p;
  p = 1;
  if (n == 0)
    return 1;
  else if (n > 0) {
    while (n > 0) {
      p = p * n;
      n--;
    }
    return p;
  }
}
int main() {
  unsigned long long int z, b, to;
  long long int n, i, c, j, d, r, h;
  c = 0, d = 0;
  unsigned long long int a[15];
  for (i = 1; i <= 15; i++) a[i] = 0;
  char s[15];
  z = 1;
  scanf("%lld", &n);
  scanf("%s", s);
  for (i = 0; i < n; i++) {
    if (s[i] == 48) c++;
    z = z * f(s[i] - 48);
  }
  j = n - 1;
  while (j >= 0) {
    if (f(s[j] - 48) % f(7) == 0) {
      a[7] = a[7] + 1;
      if (s[j] - 48 == 8) {
        a[2] = a[2] + 3;
      } else if (s[j] == 57) {
        a[3] = a[3] + 2;
        a[2] = a[2] + 3 - 2;
      }
    } else if (f(s[j] - 48) % f(5) == 0) {
      a[5] = a[5] + 1;
      if (s[j] == 54) {
        a[3] = a[3] + 1;
      }
    } else if (f(s[j] - 48) % f(3) == 0) {
      a[3] = a[3] + 1;
      if (s[j] == 52) a[2] = a[2] + 2;
    } else if (f(s[j] - 48) % f(2) == 0) {
      if (s[j] == 50) a[2] = a[2] + 1;
    }
    j--;
  }
  r = a[3] + a[2] + a[5] + a[7];
  while (a[7] > 0) {
    printf("7");
    a[7] = a[7] - 1;
  }
  while (a[5] > 0) {
    printf("5");
    a[5] = a[5] - 1;
  }
  while (a[3] > 0) {
    printf("3");
    a[3] = a[3] - 1;
  }
  while (a[2] > 0) {
    printf("2");
    a[2] = a[2] - 1;
  }
}

```