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;
class DebugStream {
} LOG;
template <typename T>
DebugStream &operator<<(DebugStream &s, const T &) {
  return s;
}
const int maxN = 5e5 + 9, maxV = 1e6 + 9, MOD = 1e9 + 7, SQ = 335, lg = 20,
          bs = 29;
int nms[6] = {4, 8, 15, 16, 23, 42};
map<int, int> ot;
int n;
vector<int> inv[6];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  for (int i = 0; i < 6; i++) ot[nms[i]] = i;
  cin >> n;
  for (int i = 0, inp; i < n; i++) {
    cin >> inp;
    inp = ot[inp];
    inv[inp].push_back(i);
  }
  auto fn = [&]() {
    for (int i = 0; i < 6; i++)
      if (inv[i].size() == 0) return true;
    return false;
  };
  for (int i = 0; i < 6; i++) reverse(inv[i].begin(), inv[i].end());
  for (int i = 0; i < 6; i++) {
    for (int el : inv[i]) LOG << el << ' ';
    LOG << '\n';
  }
  int ans = 0;
  while (!fn()) {
    int cr = 0, lm = -1;
    for (; cr < 6; cr++) {
      while (inv[cr].size() && lm > inv[cr].back()) inv[cr].pop_back();
      if (inv[cr].empty()) break;
      lm = inv[cr].back();
      inv[cr].pop_back();
    }
    if (cr == 6) ans++;
  }
  cout << n - 6 * ans << '\n';
}
