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 power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) {
      res = (res * x) % p;
    }
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long modInverse(long long a, long long p) { return power(a, p - 2, p); }
long long gcd(long long x, long long y) {
  if (x == 0 || y == 0) {
    return max(y, x);
  }
  return gcd(y % x, x);
}
int main() {
  map<long long, long long> mapa;
  mapa[4] = 1;
  mapa[8] = 2;
  mapa[15] = 3;
  mapa[16] = 4;
  mapa[23] = 5;
  mapa[42] = 6;
  long long n;
  scanf("%lld", &n);
  vector<long long> a;
  for (long long i = 0; i < n; i++) {
    long long x;
    scanf("%lld", &x);
    a.push_back(x);
  };
  vector<long long> b;
  for (auto p : a) b.push_back(mapa[p]);
  ;
  long long arr[7] = {0};
  arr[0] = n;
  long long ans = 0;
  for (auto p : b) {
    if (arr[p - 1] > 0) {
      arr[p - 1] -= 1;
      arr[p] += 1;
    }
  }
  ans = n - (6 * arr[6]);
  cout << ans << endl;
}
