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 = 500009;
int a[maxn];
queue<int> Q[7];
int main() {
  int n, good = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] == 4)
      Q[1].push(i);
    else if (a[i] == 8)
      Q[2].push(i);
    else if (a[i] == 15)
      Q[3].push(i);
    else if (a[i] == 16)
      Q[4].push(i);
    else if (a[i] == 23)
      Q[5].push(i);
    else if (a[i] == 42)
      Q[6].push(i);
  }
  while (1) {
    int x = Q[1].front();
    for (int i = 2; i <= 6; i++) {
      while (Q[i].front() < x && !Q[i].empty()) {
        Q[i].pop();
      }
      if (Q[i].empty())
        break;
      else if (i == 6 && !Q[i].empty()) {
        good += 6;
        Q[6].pop();
      } else {
        x = Q[i].front();
        Q[i].pop();
      }
    }
    if (Q[1].empty())
      break;
    else
      Q[1].pop();
  }
  cout << n - good << endl;
}
