You are given an array a consisting of n integers. Each a_i is one of the six following numbers: 4, 8, 15, 16, 23, 42.

Your task is to remove the minimum number of elements to make this array good.

An array of length k is called good if k is divisible by 6 and it is possible to split it into k/6 subsequences 4, 8, 15, 16, 23, 42.

Examples of good arrays:

  * [4, 8, 15, 16, 23, 42] (the whole array is a required sequence); 
  * [4, 8, 4, 15, 16, 8, 23, 15, 16, 42, 23, 42] (the first sequence is formed from first, second, fourth, fifth, seventh and tenth elements and the second one is formed from remaining elements); 
  * [] (the empty array is good). 



Examples of bad arrays: 

  * [4, 8, 15, 16, 42, 23] (the order of elements should be exactly 4, 8, 15, 16, 23, 42); 
  * [4, 8, 15, 16, 23, 42, 4] (the length of the array is not divisible by 6); 
  * [4, 8, 15, 16, 23, 42, 4, 8, 15, 16, 23, 23] (the first sequence can be formed from first six elements but the remaining array cannot form the required sequence). 

Input

The first line of the input contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of elements in a.

The second line of the input contains n integers a_1, a_2, ..., a_n (each a_i is one of the following numbers: 4, 8, 15, 16, 23, 42), where a_i is the i-th element of a.

Output

Print one integer — the minimum number of elements you have to remove to obtain a good array.

Examples

Input


5
4 8 15 16 23


Output


5


Input


12
4 8 4 15 16 8 23 15 16 42 23 42


Output


0


Input


15
4 8 4 8 15 16 8 16 23 15 16 4 42 23 42


Output


3
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, mn = 999999999, ans = 0;
  cin >> n;
  long long arr[n], freq[100] = {};
  map<long long, queue<long long> > mp;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    mp[arr[i]].push(i);
  }
  vector<long long> vec;
  vec.push_back(4), vec.push_back(8), vec.push_back(15), vec.push_back(16),
      vec.push_back(23), vec.push_back(42);
  while (1) {
    long long lst = -1;
    for (int i = 0; i < vec.size(); i++) {
      long long tmp = vec[i];
      if (mp[tmp].size() && mp[tmp].front() > lst) {
        lst = mp[tmp].front(), mp[tmp].pop();
      } else {
        while (mp[tmp].size() && mp[tmp].front() <= lst) {
          ans++, mp[tmp].pop();
        }
        if (!mp[tmp].size()) {
          ans += i;
          goto here;
        } else {
          lst = mp[tmp].front();
          mp[tmp].pop();
        }
      }
    }
  }
here:
  for (int i = 0; i < vec.size(); i++) ans += mp[vec[i]].size();
  cout << ans << "\n";
  return 0;
}
