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 N = 1e6 + 3;
const double PI = acos(-1);
const long long MOD = 1000000007;
long long powmod(long long a, long long b) {
  long long p = MOD;
  a %= p;
  if (!a) return 0;
  long long pr = 1;
  while (b > 0) {
    if (b & 1) {
      pr *= a;
      pr %= p;
      --b;
    }
    a *= a;
    a %= p;
    b /= 2;
  }
  return pr;
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 or n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i += 6) {
    if (n % i == 0 or n % (i + 2) == 0) return false;
  }
  return true;
}
void seive(bool a[]) {
  long long mx = sqrt(N), ii, jj;
  for (ii = 3; ii <= mx; ii += 2)
    if (!a[ii])
      for (jj = ii * ii; jj <= N; jj += 2 * ii) a[jj] = true;
}
void numofdiv(long long a[]) {
  long long mx = sqrt(N), ii, jj;
  for (ii = 1; ii <= mx; ii++) {
    for (jj = ii * ii; jj <= N; jj += ii) {
      if (jj == ii * ii)
        a[jj]++;
      else
        a[jj] += 2;
    }
  }
}
void sumofdiv(long long a[]) {
  long long mx = sqrt(N), ii, jj;
  for (ii = 1; ii <= mx; ii++) {
    for (jj = ii * ii; jj <= N; jj += ii) {
      if (jj == ii * ii)
        a[jj] += ii;
      else
        a[jj] += ii + jj / ii;
    }
  }
}
void _case() {
  long long n, i, j = 0, s = INT_MAX, ans = 0, v[6] = {0};
  cin >> n;
  long long a[n];
  for (i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 4)
      v[0]++;
    else if (a[i] == 8 and v[1] < v[0])
      v[1]++;
    else if (a[i] == 15 and v[2] < v[1])
      v[2]++;
    else if (a[i] == 16 and v[3] < v[2])
      v[3]++;
    else if (a[i] == 23 and v[4] < v[3])
      v[4]++;
    else if (a[i] == 42 and v[5] < v[4])
      v[5]++;
  }
  for (i = 0; i < 6; i++) {
    s = min(s, v[i]);
  }
  cout << n - s * 6 << '\n';
}
int main() { _case(); }
