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>
#pragma warning(disable : 4996)
using namespace std;
int n;
string a;
vector<int> v;
int main() {
  int i, j, k;
  cin >> n >> a;
  for (auto e : a) {
    if (e == '2')
      v.push_back(2);
    else if (e == '3')
      v.push_back(3);
    else if (e == '4')
      v.push_back(2), v.push_back(2), v.push_back(3);
    else if (e == '5')
      v.push_back(5);
    else if (e == '6')
      v.push_back(5), v.push_back(3);
    else if (e == '7')
      v.push_back(7);
    else if (e == '8')
      v.push_back(7), v.push_back(2), v.push_back(2), v.push_back(2);
    else if (e == '9')
      v.push_back(7), v.push_back(2), v.push_back(3), v.push_back(3);
  }
  sort(v.begin(), v.end(), greater<int>());
  for (auto e : v) cout << e;
  return 0;
}
