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 nmax = 260;
const long long inf = 1e9;
const set<long long> s = {4, 8, 15, 16, 23, 42};
vector<vector<long long>> v(6);
map<long long, long long> mp;
void initial() {
  mp[4] = 0;
  mp[8] = 1;
  mp[15] = 2;
  mp[16] = 3;
  mp[23] = 4;
  mp[42] = 5;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  initial();
  long long n;
  cin >> n;
  vector<long long> a;
  for (long long i = 0; i < n; i++) {
    long long var;
    cin >> var;
    if (s.find(var) == s.end()) continue;
    a.push_back(var);
  }
  long long cnt = 0;
  for (long long i = 0; i < a.size(); i++) {
    v[mp[a[i]]].push_back(i);
  }
  long long col = 0;
  vector<long long> pos(6, 0);
  if (v[0].size() == 0) {
    cout << n;
    return 0;
  }
  while (true) {
    for (long long i = 1; i < 6; i++) {
      long long j = pos[i];
      while (j < v[i].size() && v[i][j] < v[i - 1][pos[i - 1]]) {
        j++;
      }
      pos[i] = j;
      if (i == 5 && j != v[i].size()) col++;
      if (j == v[i].size()) goto laza;
    }
    for (long long i = 0; i < 6; i++) {
      pos[i]++;
    }
  }
laza:
  cout << n - col * 6;
  return 0;
}
