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;
long long q, dp[105][100005], a[555555], b[555555], k, l, m, n, o, p;
map<long long, long long> mp;
vector<long long> adj[555555];
const long long mod = 1e9 + 7;
void solve() {
  long long c[] = {4, 8, 15, 16, 23, 42};
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    for (long long j = 0; j < 6; j++) {
      if (a[i] == c[j]) {
        a[i] = j;
        break;
      }
    }
  }
  for (long long i = 0; i < n; i++) {
    if (!a[i])
      b[0]++;
    else {
      if (b[a[i] - 1]) {
        b[a[i] - 1]--;
        b[a[i]]++;
      }
    }
  }
  cout << n - b[5] * 6 << endl;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  q = 1;
  while (q--) {
    solve();
  }
}
