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;
vector<int> tekijat[10] = {{},
                           {},
                           {2},
                           {2, 3},
                           {2, 3, 2, 2},
                           {2, 3, 2, 2, 5},
                           {2, 3, 2, 2, 5, 2, 3},
                           {2, 3, 2, 2, 5, 2, 3, 7},
                           {2, 3, 2, 2, 5, 2, 3, 7, 2, 2, 2},
                           {2, 3, 2, 2, 5, 2, 3, 7, 2, 2, 2, 3, 3}};
int main() {
  vector<int> o(10);
  int n;
  string a;
  cin >> n >> a;
  for (char c : a) {
    int v = c - '0';
    for (int t : tekijat[v]) o[t]++;
  }
  vector<int> t;
  while (o[7] >= 1 && o[5] >= 1 && o[3] >= 2 && o[2] >= 4) {
    t.push_back(7);
    o[7]--;
    o[5]--;
    o[3] -= 2;
    o[2] -= 4;
  }
  while (o[5] >= 1 && o[3] >= 1 && o[2] >= 3) {
    t.push_back(5);
    o[5]--;
    o[3]--;
    o[2] -= 3;
  }
  while (o[3] >= 1 && o[2] >= 1) {
    t.push_back(3);
    o[3]--;
    o[2]--;
  }
  while (o[2] >= 1) {
    t.push_back(2);
    o[2]--;
  }
  sort(t.rbegin(), t.rend());
  for (int v : t) cout << v;
  cout << '\n';
}
