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;
int biao[10];
long long n, a;
vector<int> ans;
void Load() {
  biao[1] = 1;
  for (int i = 2; i < 10; i++) biao[i] = biao[i - 1] * i;
  cin >> n >> a;
}
int main() {
  Load();
  while (a) {
    int now = a % 10;
    int out = biao[now];
    while (out) {
      if (out == 1) break;
      if (out % biao[7] == 0) {
        out /= biao[7];
        ans.push_back(7);
      }
      if (out % biao[5] == 0) {
        out /= biao[5];
        ans.push_back(5);
      }
      if (out % biao[3] == 0) {
        out /= biao[3];
        ans.push_back(3);
      }
      if (out % 2 == 0) {
        out /= 2;
        ans.push_back(2);
      }
    }
    a /= 10;
  }
  sort(ans.begin(), ans.end());
  for (int i = ans.size() - 1; i >= 0; i--) cout << ans[i];
  cout << endl;
  return 0;
}
