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;
bool comp(const pair<long long, long long> &a,
          const pair<long long, long long> &b) {
  return a.second < b.second;
}
int main() {
  long long n, m = 100001;
  cin >> n;
  long long a[n], p[100] = {0}, c[6] = {4, 8, 15, 16, 23, 42};
  for (long long i = 0; i < n; i++) cin >> a[i];
  if (n < 6)
    cout << n;
  else {
    for (long long i = n - 1; i >= 0; i--) {
      if (a[i] == 42) p[42]++;
      if (a[i] == 23 && p[23] + 1 <= p[42]) p[23]++;
      if (a[i] == 16 && p[16] + 1 <= p[23]) p[16]++;
      if (a[i] == 15 && p[15] + 1 <= p[16]) p[15]++;
      if (a[i] == 8 && p[8] + 1 <= p[15]) p[8]++;
      if (a[i] == 4 && p[4] + 1 <= p[8]) p[4]++;
    }
    for (long long i = 0; i < 6; i++) m = min(m, p[c[i]]);
    cout << n - m * 6;
  }
}
