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 factorial(int n) { return (n == 0) ? 1 : n * factorial(n - 1); }
int main() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  int counter2 = 0, counter3 = 0, counter5 = 0, counter7 = 0;
  for (int i = 0; i < n; i++) {
    char x = s[i];
    if (x == '5') {
      counter5++;
    }
    if (x == '6') {
      counter5++;
      counter3++;
    }
    if (x == '7') {
      counter7++;
    }
    if (x == '8') {
      counter2 += 3;
      counter7++;
    }
    if (x >= '9') {
      counter7++;
      counter3 += 2;
      counter2++;
    }
    if (x == '3') {
      counter3++;
    }
    if (x == '4') {
      counter3++;
      counter2 += 2;
    }
    if (x == '2') {
      counter2++;
    }
  }
  for (int i = 0; i < counter7; i++) cout << "7";
  for (int i = 0; i < counter5; i++) cout << "5";
  for (int i = 0; i < counter3; i++) cout << "3";
  for (int i = 0; i < counter2; i++) cout << "2";
  cout << '\n';
}
