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;
const int inf = 1e9 + 5;
const int N = 5e5 + 5;
const int mod = 1e9 + 7;
const int Mx = 1e7;
vector<int> pos[N];
int a[6] = {4, 8, 15, 16, 23, 42};
int New[100];
int vis[N];
void solve() {
  int n;
  cin >> n;
  int x;
  for (int i = 0; i < 50; i++) pos[i].clear();
  for (int i = 0; i < n; i++) {
    cin >> x;
    pos[x].push_back(i);
  }
  int Mi = inf;
  for (int i = 0; i < 6; i++) {
    Mi = min(int(pos[a[i]].size()), Mi);
  }
  for (int i = 0; i < 6; i++) {
    pos[a[i]].push_back(-1);
  }
  int cnt;
  bool f = true;
  for (cnt = 0; cnt <= Mi; cnt++) {
    int last = -1;
    for (int i = 0; i < 6; i++) {
      for (int j = New[a[i]]; j < pos[a[i]].size(); j++) {
        int tmp = pos[a[i]][j];
        if (tmp == -1) {
          f = false;
          break;
        }
        if (vis[tmp])
          continue;
        else {
          if (last < tmp) {
            vis[tmp] = true;
            last = tmp;
            New[a[i]] = j + 1;
            break;
          } else
            continue;
        }
      }
      if (!f) break;
    }
    if (!f) break;
  }
  cout << n - cnt * 6 << endl;
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
