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 Max = 2e6 + 10;
const int Mod = 1e9 + 7;
const long long int Inf = 1LL << 62;
set<int> pos[Max];
int ar[Max];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &ar[i]);
  }
  int tot = 0;
  for (int i = 1; i <= n; i++) {
    pos[ar[i]].insert(i);
    if (pos[4].size() == 0) {
      continue;
    }
    int lst = *pos[4].begin();
    bool f = 0;
    for (int x : {4, 8, 15, 16, 23, 42}) {
      if (!pos[x].size()) {
        f = 1;
        break;
      }
      int prv = *pos[x].lower_bound(lst);
      if (prv < lst || pos[x].find(prv) == pos[x].end()) {
        f = 1;
        break;
      }
      lst = prv;
    }
    if (f) {
      continue;
    }
    tot += 1;
    lst = *pos[4].begin();
    for (int x : {4, 8, 15, 16, 23, 42}) {
      lst = *pos[x].lower_bound(lst);
      pos[x].erase(*pos[x].lower_bound(lst));
    }
  }
  printf("%d\n", n - 6 * tot);
  return 0;
}
