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 GCC optimize("O3")
#pragma GCC target("sse4")
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long power(long long a, long long b, long long m = 1000000007) {
  long long answer = 1;
  while (b) {
    if (b & 1) answer = (answer * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return answer;
}
using namespace std;
void ctrl() { cout << "Control" << endl; }
long long make_num(string p) {
  stringstream geek(p);
  long long x = 0;
  geek >> x;
  return x;
}
string make_str(long long x) {
  ostringstream str1;
  str1 << x;
  string geek = str1.str();
  return geek;
}
string get(char x) {
  if (x == '2') return "2";
  if (x == '3') return "3";
  if (x == '4') return "322";
  if (x == '5') return "5";
  if (x == '6') return "53";
  if (x == '7') return "7";
  if (x == '8') return "7222";
  if (x == '9') return "7332";
  return "";
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  long long n;
  cin >> n;
  string p;
  cin >> p;
  string x = "";
  for (long long i = 0; i < p.length(); i++) {
    x += get(p[i]);
  }
  sort(x.begin(), x.end());
  reverse(x.begin(), x.end());
  cout << x;
  return 0;
}
