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;
long long binpow(long long a, long long b) {
  a %= 1000000007;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
    b >>= 1;
  }
  return res;
}
void fun(int x, vector<int>& v) {
  if (x == 4) {
    v.push_back(3);
    v.push_back(2);
    v.push_back(2);
  }
  if (x == 6) {
    v.push_back(5);
    v.push_back(3);
  }
  if (x == 8) {
    v.push_back(7);
    v.push_back(2);
    v.push_back(2);
    v.push_back(2);
  }
  if (x == 9) {
    v.push_back(3);
    v.push_back(3);
    v.push_back(7);
    v.push_back(2);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  string s;
  cin >> s;
  int a[n];
  for (long long i = 0; i < n; i++) {
    a[i] = s[i] - 48;
  }
  vector<int> v;
  for (long long i = 0; i < n; i++) {
    if (a[i] == 4 || a[i] == 6 || a[i] == 8 || a[i] == 9) {
      fun(a[i], v);
    } else
      v.push_back(a[i]);
  }
  sort(v.begin(), v.end(), greater<int>());
  for (long long i = 0; i < v.size(); i++) {
    if (v[i] < 2) break;
    cout << v[i];
  }
}
