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;
string ConvertToBin(long long s) {
  string res = "";
  while (s != 0) {
    res += (char)('0' + s % 2);
    s /= 2;
  }
  reverse(res.begin(), res.end());
  return res;
}
void PrintVec(vector<int> &v) {
  for (vector<int>::iterator it = v.begin(); it != v.end(); ++it)
    cout << *it << " ";
}
vector<int> Populate(vector<int> &v, int n) {
  int num;
  for (int i = 0; i < n; i++) {
    cin >> num;
    v.push_back(num);
  }
  return (v);
}
int sumofdigits(int n) {
  int sm = 0;
  while (n) {
    sm = sm + (n - (n / 10) * 10);
    n = n / 10;
  }
  return (sm);
}
bool isprime(int n) {
  int i = 0;
  for (i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return (false);
  }
  return (true);
}
int main() {
  int n, i, j, count = 0, count2 = 0, min = 1000000;
  cin >> n;
  int arr[n];
  for (i = 0; i < n; i++) {
    cin >> arr[i];
  }
  int ar[43];
  for (i = 0; i < 43; i++) {
    if (i == 4 || i == 8 || i == 15 || i == 16 || i == 23 || i == 42)
      ar[i] = 0;
    else
      ar[i] = 1000000;
  }
  for (i = 0; i < n; i++) {
    ar[arr[i]]++;
    for (j = 3; j <= arr[i] - 1; j++) {
      if (ar[j] < ar[arr[i]]) {
        ar[arr[i]]--;
        count++;
      }
    }
  }
  for (i = 0; i < 43; i++) {
    if (ar[i] < min) min = ar[i];
  }
  for (i = 0; i < 43; i++) {
    if (ar[i] != 1000000) count2 = count2 + (ar[i] - min);
  }
  cout << count + count2;
  return (0);
}
