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>
const int inf = 1000005;
using namespace std;
int dx[] = {1, 1, 1, 0, 0, -1, -1, -1};
int dy[] = {1, 0, -1, 1, -1, -1, 0, 1};
vector<int> arr[10];
void compute() {
  int temp[] = {2, 3, 5, 7};
  for (int i = 1; i < 10; ++i) {
    int num = i;
    for (int j = 0; j < 4; ++j) {
      while (num % temp[j] == 0) {
        arr[i].push_back(temp[j]);
        num /= temp[j];
      }
    }
  }
}
int cnt[10];
void add(int n) {
  for (int i = 2; i <= n; ++i) {
    for (int x : arr[i]) cnt[x]++;
  }
}
void rmv(int n) {
  for (int i = 2; i <= n; ++i) {
    for (int x : arr[i]) cnt[x]--;
  }
}
string func(string &str, int n) {
  string ans;
  for (char c : str) {
    if (c == '0' || c == '1') continue;
    add(c - '0');
  }
  int i = 9;
  while (i >= 2) {
    if (!cnt[i]) {
      i--;
      continue;
    }
    ans.push_back(i + '0');
    rmv(i);
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  compute();
  int n;
  cin >> n;
  string str;
  cin >> str;
  cout << func(str, n) << "\n";
  return 0;
}
