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;
int n;
string a;
char x[250];
int main() {
  cin >> n >> a;
  int l = 0;
  for (int i = 0; i < a.size(); i++) {
    int d = a[i] - '0';
    if (a[i] == '0' || a[i] == '1') continue;
    if (a[i] == '2' || a[i] == '3' || a[i] == '5' || a[i] == '7')
      x[l++] += a[i];
    else {
      if (d == 4) {
        x[l++] += '3';
        x[l++] += '2';
        x[l++] += '2';
      }
      if (d == 6) {
        x[l++] += '5';
        x[l++] += '3';
      };
      if (d == 8) {
        x[l++] += '7';
        x[l++] += '2';
        x[l++] += '2';
        x[l++] += '2';
      }
      if (d == 9) {
        x[l++] += '7';
        x[l++] += '3';
        x[l++] += '3';
        x[l++] += '2';
      }
    }
  }
  sort(x, x + l);
  reverse(x, x + l);
  cout << x << endl;
  return 0;
}
