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 maxn = 500100;
const int INF = (1LL << 29);
int n, a[maxn];
int v[6];
int b[6] = {4, 8, 15, 16, 23, 42};
int c[6] = {1, 2, 3, 4, 5, 0};
int id(int x) {
  for (int i = 0; i <= 5; i++)
    if (b[i] == x) return i;
  return -1;
}
void solve() {
  for (int i = 1; i <= n; i++) cin >> a[i];
  memset(v, 0, sizeof(v));
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == 4)
      v[0]++;
    else {
      int x = id(a[i]);
      if (x == -1)
        cnt++;
      else if (v[x - 1] == 0)
        cnt++;
      else
        v[x - 1]--, v[x]++;
    }
  }
  int res = cnt;
  for (int i = 0; i <= 4; i++) res += v[i] * c[i];
  cout << res << endl;
}
int main() {
  while (cin >> n) solve();
  return 0;
}
