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>
using namespace std;
char str[16];
int num[10];
int main() {
  int n;
  while (~scanf("%d", &n)) {
    scanf("%s", str);
    memset(num, 0, sizeof(num));
    for (int i = 0; i < n; ++i) {
      if (str[i] == '0' || str[i] == '1')
        continue;
      else if (str[i] == '2' || str[i] == '3' || str[i] == '5' || str[i] == '7')
        num[str[i] - '0']++;
      else if (str[i] == '4') {
        num[2] = num[2] + 2;
        num[3]++;
      } else if (str[i] == '6') {
        num[5]++;
        num[3]++;
      } else if (str[i] == '8') {
        num[7]++;
        num[2] = num[2] + 3;
      } else {
        num[7]++;
        num[3] = num[3] + 2;
        num[2]++;
      }
    }
    for (int i = 9; i >= 2; --i) {
      for (int j = 1; j <= num[i]; ++j) {
        printf("%d", i);
      }
    }
    printf("\n");
  }
  return 0;
}
