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;
const int N = 1e5 + 5;
const int MAX = 1e9 + 5;
int n;
char c;
vector<int> v;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf(" %c", &c);
    if (c == '4')
      v.push_back(2), v.push_back(2), v.push_back(3);
    else if (c == '6')
      v.push_back(5), v.push_back(3);
    else if (c == '8')
      v.push_back(2), v.push_back(2), v.push_back(2), v.push_back(7);
    else if (c == '9')
      v.push_back(3), v.push_back(3), v.push_back(2), v.push_back(7);
    else if (c > '1')
      v.push_back(c - '0');
  }
  sort(v.begin(), v.end(), greater<int>());
  for (__typeof(v.begin()) it = v.begin(); it != v.end(); it++) cout << *it;
  cout << '\n';
  return 0;
}
