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;
signed long long int logo(signed long long int n) {
  if (n == 1) return 0;
  return 1 + logo(n / 2);
}
signed long long int fn(signed long long int n) {
  signed long long int ans = 0;
  while (n % 3 == 0) {
    n *= 2;
    n /= 3;
    ans++;
  }
  while (n % 5 == 0) {
    n = (n * 4) / 5;
    ans++;
  }
  if (n & (n - 1)) return -1;
  return ans + logo(n);
}
void solve() {
  int n, o = 0, t = 0, ans = 0;
  cin >> n;
  int a[n];
  int f[7] = {0};
  f[0] = INT_MAX;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 4)
      f[1]++;
    else {
      if (a[i] == 42)
        f[6] = min(f[6] + 1, f[5]);
      else
        f[a[i] / 4] = min(f[a[i] / 4 - 1], f[a[i] / 4] + 1);
    }
  }
  cout << n - f[6] * 6;
}
int main() {
  int t = 1;
  while (t--) solve();
  return 0;
}
