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;
int main() {
  long long int n;
  cin >> n;
  long long int i, x, j;
  long long int b[6] = {4, 8, 15, 16, 23, 42}, a[n];
  vector<long long int> v[45];
  for (i = 0; i < n; i++) {
    cin >> a[i];
    v[a[i]].push_back(i);
  }
  long long int c = 0;
  long long int vp[6] = {0};
  for (i = 0; i < v[b[0]].size(); i++) {
    long long int pos = v[b[0]][i];
    long long int f = 0;
    for (j = 1; j < 6; j++) {
      long long int m = v[b[j]].size(), k;
      for (k = vp[j]; k < m; k++)
        if (v[b[j]][k] > pos) break;
      if (k == m) {
        f = 1;
        break;
      } else {
        pos = v[b[j]][k];
        vp[j] = k + 1;
      }
    }
    if (f == 0) c++;
    if (f) break;
  }
  cout << n - 6 * c;
  return 0;
}
