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;
int v[100];
int main() {
  int x, n;
  scanf("%d", &n);
  int ans = 0, k = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &x);
    if (x == 4) {
      v[4]++;
    } else if (x == 8) {
      if (v[4] > 0) {
        v[4]--;
        v[8]++;
      } else
        k++;
    } else if (x == 15) {
      if (v[8] > 0) {
        v[8]--;
        v[15]++;
      } else
        k++;
    } else if (x == 16) {
      if (v[15] > 0) {
        v[15]--;
        v[16]++;
      } else
        k++;
    } else if (x == 23) {
      if (v[16] > 0) {
        v[16]--;
        v[23]++;
      } else
        k++;
    } else if (x == 42) {
      if (v[23] > 0) {
        v[23]--;
        v[42]++;
      } else
        k++;
    }
  }
  k += (v[4]) + (2 * v[8]) + (3 * v[15]) + (4 * v[16]) + (5 * v[23]);
  cout << k << endl;
  return 0;
}
