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() {
  int n;
  scanf("%d", &n);
  vector<int> v(6, 0);
  int x;
  int four = 0;
  int eight = 0;
  int fif = 0;
  int six = 0;
  int twen = 0;
  int forty = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &x);
    if (x == 4) four = i, v[0] += 1;
    if (x == 8 && v[0] > 0 && i > four) eight = i, v[1] += 1, v[0] -= 1;
    if (x == 15 && v[1] > 0 && i > eight) fif = i, v[2] += 1, v[1] -= 1;
    if (x == 16 && v[2] > 0 && i > fif) six = i, v[3] += 1, v[2] -= 1;
    if (x == 23 && v[3] > 0 && i > six) twen = i, v[4] += 1, v[3] -= 1;
    if (x == 42 && v[4] > 0 && i > twen) forty = i, v[5] += 1, v[4] -= 1;
  }
  if (v[5] <= 0)
    cout << n << endl;
  else
    cout << n - (v[5] * 6) << endl;
}
