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 long long inf = 1e18;
const int N = 2e6 + 10;
const long long mod = 1e9 + 7;
const int base = 131;
const double pi = acos(-1);
const double eps = 1e-8;
long long vis[N], num[N], a[N], n, m, k, x, y, z;
long long cx, cy, cnt, ans, sum, flag, t, ff;
map<pair<int, int>, int> mp;
set<int> sa, sb;
vector<pair<int, int> > p;
vector<int> res;
string s;
stack<long long> st;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) {
    if (a[i] == 4)
      num[4]++;
    else if (a[i] == 8 && num[4]) {
      num[4]--;
      num[8]++;
    } else if (a[i] == 15 && num[8]) {
      num[8]--;
      num[15]++;
    } else if (a[i] == 16 && num[15])
      num[15]--, num[16]++;
    else if (a[i] == 23 && num[16])
      num[23]++, num[16]--;
    else if (a[i] == 42 && num[23])
      num[42]++, num[23]--;
  }
  cout << n - num[42] * 6 << endl;
}
