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;
struct node {
  long long x;
  long long y;
};
long long gcd(long long a, long long b) {
  return b == 0 ? a : a > b ? gcd(b, a % b) : gcd(a, b % a);
}
bool myfunc(struct node a, struct node b) {
  if (a.x == b.x) {
    return a.y < b.y;
  } else {
    return a.x < b.x;
  }
}
string s;
vector<string> v;
vector<string> arr[10];
int main() {
  int n;
  string temp = "";
  cin >> n;
  cin >> s;
  int i;
  for (i = 0; i < n; i++) {
    if (s[i] == '4') {
      v.push_back("2");
      v.push_back("2");
      v.push_back("3");
    } else if (s[i] == '6') {
      v.push_back("5");
      v.push_back("3");
    } else if (s[i] == '8') {
      v.push_back("2");
      v.push_back("2");
      v.push_back("7");
      v.push_back("2");
    } else if (s[i] == '9') {
      v.push_back("7");
      v.push_back("3");
      v.push_back("3");
      v.push_back("2");
    } else {
      int j;
      string temp = "";
      temp = temp + s[i];
      v.push_back(temp);
    }
  }
  sort(v.rbegin(), v.rend());
  string ans = "";
  for (i = 0; i < v.size(); i++) {
    if (v[i] == "0" || v[i] == "1") {
      break;
    } else {
      ans += v[i];
    }
  }
  cout << ans << endl;
  return 0;
}
