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 N = 5 * 1e5 + 10;
int n, a[N];
int s[] = {4, 8, 15, 16, 23, 42};
int counts[50];
int mu[7];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  if (n < 6) {
    cout << n << endl;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] == s[0]) {
      mu[1]++;
    } else if (a[i] == s[1] && mu[1] > 0 && mu[1] > mu[2]) {
      mu[2]++;
    } else if (a[i] == s[2] && mu[2] > 0 && mu[2] > mu[3]) {
      mu[3]++;
    } else if (a[i] == s[3] && mu[3] > 0 && mu[3] > mu[4]) {
      mu[4]++;
    } else if (a[i] == s[4] && mu[4] > 0 && mu[4] > mu[5]) {
      mu[5]++;
    } else if (a[i] == s[5] && mu[5] > 0 && mu[5] > mu[6]) {
      mu[6]++;
    }
  }
  int cnt = mu[1];
  for (int i = 1; i <= 6; i++) {
    cnt = min(cnt, mu[i]);
  }
  cout << (n - 6 * cnt) << endl;
}
