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;
const int MOD = 1e9 + 7;
const int N = 15;
int cnt[10];
vector<int> ans;
int main() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '2') cnt[2]++;
    if (s[i] == '3') {
      cnt[2]++;
      cnt[3]++;
    }
    if (s[i] == '4') {
      cnt[2] += 3;
      cnt[3]++;
    }
    if (s[i] == '5') {
      cnt[2] += 3;
      cnt[3]++;
      cnt[5]++;
    }
    if (s[i] == '6') {
      cnt[2] += 4;
      cnt[3] += 2;
      cnt[5]++;
    }
    if (s[i] == '7') {
      cnt[2] += 4;
      cnt[3] += 2;
      cnt[5]++;
      cnt[7]++;
    }
    if (s[i] == '8') {
      cnt[2] += 7;
      cnt[3] += 2;
      cnt[5]++;
      cnt[7]++;
    }
    if (s[i] == '9') {
      cnt[2] += 7;
      cnt[3] += 4;
      cnt[5]++;
      cnt[7]++;
    }
  }
  if (cnt[7]) {
    cnt[5] = cnt[5] - (cnt[7] * 1);
    cnt[3] = cnt[3] - (cnt[7] * 2);
    cnt[2] = cnt[2] - (cnt[7] * 4);
    for (int i = 0; i < cnt[7]; i++) ans.push_back(7);
  }
  if (cnt[5]) {
    cnt[3] = cnt[3] - (cnt[5] * 1);
    cnt[2] = cnt[2] - (cnt[5] * 3);
    for (int i = 0; i < cnt[5]; i++) ans.push_back(5);
  }
  if (cnt[3]) {
    cnt[2] = cnt[2] - (cnt[3] * 1);
    for (int i = 0; i < cnt[3]; i++) ans.push_back(3);
  }
  if (cnt[2]) {
    for (int i = 0; i < cnt[2]; i++) ans.push_back(2);
  }
  sort(ans.begin(), ans.end());
  reverse(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) cout << ans[i];
  return 0;
}
