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 n;
char str[20];
int cnt[20];
int main() {
  vector<int> G[10];
  G[2].push_back(2);
  G[3].push_back(3);
  G[4].push_back(2);
  G[4].push_back(2);
  G[5].push_back(5);
  G[6].push_back(2);
  G[6].push_back(3);
  G[7].push_back(7);
  G[8].push_back(2);
  G[8].push_back(2);
  G[8].push_back(2);
  G[9].push_back(3);
  G[9].push_back(3);
  int a[] = {2, 3, 5, 7};
  while (scanf("%d", &n) == 1) {
    scanf("%s", str);
    memset(cnt, 0, sizeof(cnt));
    vector<int> ans;
    for (int i = 0; str[i]; i++)
      for (int j = str[i] - '0'; j >= 0; j--)
        for (int k = 0; k < G[j].size(); k++) cnt[G[j][k]]++;
    for (int i = 3; i >= 0; i--)
      while (cnt[a[i]]) {
        for (int j = a[i]; j >= 0; j--)
          for (int k = 0; k < G[j].size(); k++) cnt[G[j][k]]--;
        ans.push_back(a[i]);
      }
    sort(ans.begin(), ans.end());
    reverse(ans.begin(), ans.end());
    for (int i = 0; i < ans.size(); i++) printf("%d", ans[i]);
    printf("\n");
  }
  return 0;
}
