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 BUFFER[100000 + 5];
bool readn(int &n) { return scanf("%d", &n) == 1; }
bool readl(long long &n) { return scanf("%I64d", &n) == 1; }
bool readd(double &n) { return scanf("%lf", &n) == 1; }
bool reads(string &s) {
  s = "";
  int n = scanf("%s", BUFFER);
  if (n == 1) s = BUFFER;
  return n == 1;
}
bool readln(string &s) {
  char *valid = gets(BUFFER);
  if (valid) s = BUFFER;
  return ((bool)valid);
}
string x[10] = {"", "", "2", "3", "322", "5", "53", "7", "7222", "7332"};
int main() {
  int i, n;
  string s, ans = "";
  cin >> n >> s;
  for (i = 0; i < ((int)s.size()); ++i) ans += x[s[i] - '0'];
  sort(ans.begin(), ans.end());
  reverse(ans.begin(), ans.end());
  cout << ans << endl;
  return 0;
}
