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;
template <class T>
void show(const vector<T> &a) {
  for (T x : a) cout << x << " ";
  cout << '\n';
}
const long long sze = 5e5 + 50, oo = 1e18 + 500, mod = 1e9 + 7;
const long double eps = 1e-9, PI = 2 * acos(0.0);
vector<long long> vertices[sze];
vector<long long> visit(sze, 0);
vector<char> used(sze, false);
vector<long long> arr(sze, 0);
long long n, m, k;
long long A, B;
long long cnt = 0;
signed main() {
  ios::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  vector<long long> vec = {4, 8, 15, 16, 23, 42};
  cin >> n;
  vector<long long> a(n);
  vector<vector<long long> > ind(6);
  vector<long long> pos(6, 0);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    for (long long j = 0; j < 6; j++) {
      if (a[i] == vec[j]) {
        ind[j].push_back(i);
      }
    }
  }
  cnt = 0;
  while (1) {
    if (pos[0] >= ((long long)(ind[0].size()))) break;
    long long i = ind[0][pos[0]];
    pos[0]++;
    for (long long j = 1; j < 6; j++) {
      bool fl = false;
      for (; pos[j] < ((long long)(ind[j].size())); pos[j]++) {
        if (ind[j][pos[j]] > i) {
          fl = true;
          i = ind[j][pos[j]];
          pos[j]++;
          break;
        }
      }
      if (!fl) {
        return cout << n - cnt * 6, 0;
      }
    }
    cnt++;
  }
  cout << n - cnt * 6;
  return 0;
}
