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 MAXN = 55;
int a[MAXN];
int nxt[6];
long long solve() {
  int n;
  scanf("%d", &n);
  vector<int> ids[6];
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    if (x == 4) ids[0].push_back(i);
    if (x == 8) ids[1].push_back(i);
    if (x == 15) ids[2].push_back(i);
    if (x == 16) ids[3].push_back(i);
    if (x == 23) ids[4].push_back(i);
    if (x == 42) ids[5].push_back(i);
  }
  for (int i = 0; i < 6; i++) nxt[i] = 0;
  int res = 0;
  for (int i : ids[0]) {
    int lst = i;
    nxt[0]++;
    for (int j = 1; j < 6; j++) {
      while (nxt[j] < ids[j].size() && ids[j][nxt[j]] < lst) {
        nxt[j]++;
        res++;
      }
      if (nxt[j] == ids[j].size()) {
        for (int k = 0; k < 6; k++) {
          res += ids[k].size() - nxt[k];
        }
        res += j;
        return res;
      }
      lst = ids[j][nxt[j]];
      nxt[j]++;
    }
  }
  int len = ids[0].size();
  return (n - res) - len * 6;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T;
  T = 1;
  while (T--) {
    printf("%lld\n", solve());
  }
  return 0;
}
