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;
bool check(long long *t1, long long *t2, long long *t3, long long *t4,
           long long *t5, long long *t6, long long *c) {
  if (*t1 < *t2) {
    (*t2)--;
    (*c)++;
  } else if (*t2 < *t3) {
    (*t3)--;
    (*c)++;
  } else if (*t3 < *t4) {
    (*t4)--;
    (*c)++;
  } else if (*t4 < *t5) {
    (*t5)--;
    (*c)++;
  } else if (*t5 < *t6) {
    (*t6)--;
    (*c)++;
  } else
    return true;
  return false;
}
long long find_min(long long t1, long long t2, long long t3, long long t4,
                   long long t5, long long t6) {
  long long min = INT_MAX;
  if (t1 < min) {
    min = t1;
  }
  if (t2 < min) {
    min = t2;
  }
  if (t3 < min) {
    min = t3;
  }
  if (t4 < min) {
    min = t4;
  }
  if (t5 < min) {
    min = t5;
  }
  if (t6 < min) {
    min = t6;
  }
  return min;
}
int main() {
  long long n, num, t1, t2, t3, t4, t5, t6, c = 0;
  cin >> n;
  vector<long long> v;
  t1 = t2 = t3 = t4 = t5 = t6 = 0;
  for (long long i = 0; i < n; ++i) {
    cin >> num;
    if (num == 4)
      t1++;
    else if (num == 8)
      t2++;
    else if (num == 15)
      t3++;
    else if (num == 16)
      t4++;
    else if (num == 23)
      t5++;
    else if (num == 42)
      t6++;
    else
      c++;
    if (check(&t1, &t2, &t3, &t4, &t5, &t6, &c)) {
      v.push_back(num);
    }
  }
  if (t1 == t2 && t2 == t3 && t3 == t4 && t4 == t5 && t5 == t6) {
    cout << c;
    return 0;
  }
  long long min = find_min(t1, t2, t3, t4, t5, t6);
  c += (v.size() - min * 6);
  cout << c;
  return 0;
}
