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 NMAX = 1000005;
int v[NMAX];
int a[] = {4, 8, 15, 16, 23, 42};
int st[10];
int findpoz(int x) {
  int i = 0;
  while (x != a[i]) {
    i++;
  }
  return i;
}
int main() {
  int n;
  scanf("%d", &n);
  bool ok = 0;
  int p = 0;
  int sol = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &v[++p]);
    int j;
    for (j = 0; j < 6; j++) {
      if (v[i] == a[j]) {
        j = 10;
      }
    }
    if (j < 10) {
      v[p--] = 0;
      sol++;
    }
  }
  n = p;
  for (int i = 1; i <= n; i++) {
    int poz = findpoz(v[i]) + 1;
    if (st[poz - 1] == 0) {
      if (poz == 1) {
        st[poz]++;
      } else {
        sol++;
      }
    } else {
      st[poz - 1]--;
      st[poz]++;
    }
  }
  for (int j = 1; j <= 5; j++) {
    sol += (st[j]) * j;
  }
  printf("%d\n", sol);
  return 0;
}
