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;
using namespace std;
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  set<long long> s[6];
  for (long long i = 0; i < n; i++) {
    if (a[i] == 4) s[0].insert(i);
    if (a[i] == 8) s[1].insert(i);
    if (a[i] == 15) s[2].insert(i);
    if (a[i] == 16) s[3].insert(i);
    if (a[i] == 23) s[4].insert(i);
    if (a[i] == 42) s[5].insert(i);
  }
  long long ans = 0, fla = 0;
  while (fla == 0) {
    if (s[0].empty()) break;
    long long i = *s[0].begin();
    s[0].erase(i);
    long long j = 1;
    while (j < 6) {
      set<long long>::iterator p = s[j].upper_bound(i);
      if (p == s[j].end()) {
        fla = 1;
        break;
      } else {
        i = *p;
        s[j].erase(*p);
        j++;
      }
    }
    if (fla == 1) break;
    if (j == 6) ans++;
  }
  cout << n - 6 * ans << "\n";
}
int main() {
  ios::sync_with_stdio(true);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
