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, class... U>
void pr(T&& a, U&&... b) {
  cout << forward<T>(a);
  (int[]){(cout << forward<U>(b), 0)...};
}
template <class T, class... U>
void re(T&& a, U&&... b) {
  cin >> forward<T>(a);
  (int[]){(cin >> forward<U>(b), 0)...};
}
const int MAXN = 5e5;
const int INF = 1e9;
const int MOD = 1e9 + 7;
int N, A[MAXN + 5];
int cnt1, cnt2, cnt3, cnt4, cnt5, cnt6;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  re(N);
  for (int i = 1; i <= N; i++) re(A[i]);
  for (int i = 1; i <= N; i++) switch (A[i]) {
      case 4:
        cnt1++;
        break;
      case 8:
        if (cnt1 > 0) cnt2++, cnt1--;
        break;
      case 15:
        if (cnt2 > 0) cnt3++, cnt2--;
        break;
      case 16:
        if (cnt3 > 0) cnt4++, cnt3--;
        break;
      case 23:
        if (cnt4 > 0) cnt5++, cnt4--;
        break;
      case 42:
        if (cnt5 > 0) cnt6++, cnt5--;
        break;
    }
  pr(N - 6 * cnt6);
}
