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>
#include <bits/stdc++.h>
int main() {
  int n, temp;
  scanf("%d", &n);
  char s[20];
  scanf("%s", s);
  int ans[2000];
  int z = 0;
  memset(ans, 0, sizeof(ans));
  for (int i = 0; i < n; i++) {
    int j = s[i] - '0';
    if (j == 9) {
      ans[z++] = 7;
      ans[z++] = 3;
      ans[z++] = 3;
      ans[z++] = 2;
    } else if (j == 8) {
      ans[z++] = 7;
      ans[z++] = 2;
      ans[z++] = 2;
      ans[z++] = 2;
    } else if (j == 6) {
      ans[z++] = 5;
      ans[z++] = 3;
    } else if (j == 4) {
      ans[z++] = 3;
      ans[z++] = 2;
      ans[z++] = 2;
    } else if (j != 1 && j != 0)
      ans[z++] = j;
  }
  int x = 0;
  for (int i = 0; ans[i] != 0; i++) x++;
  for (int i = 0; i < x - 1; i++)
    for (int j = i + 1; j < x; j++)
      if (ans[i] < ans[j]) {
        temp = ans[i];
        ans[i] = ans[j];
        ans[j] = temp;
      }
  for (int i = 0; i < x; i++) printf("%d", ans[i]);
}
