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;
void sol() {
  long long int n;
  cin >> n;
  vector<int> ar[6];
  long long int a;
  vector<int> v = {4, 8, 15, 16, 23, 42};
  for (long long int i = 0; i < n; i++) {
    cin >> a;
    ar[lower_bound(v.begin(), v.end(), a) - v.begin()].push_back(i);
  }
  int p[6] = {0, 0, 0, 0, 0, 0};
  int sit = 0;
  int cnt = 0;
  while (1) {
    int prev = -1;
    for (long long int i = 0; i < 6; i++) {
      while (p[i] < ar[i].size() && ar[i][p[i]] < prev) {
        p[i]++;
      }
      if (p[i] >= ar[i].size()) {
        sit = 1;
        break;
      }
      prev = ar[i][p[i]];
      p[i]++;
    }
    if (sit == 1)
      break;
    else
      cnt++;
  }
  cout << n - 6 * cnt << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    sol();
  }
  return 0;
}
