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 main() {
  long long n;
  scanf("%lld", &n);
  vector<long long> input(n);
  for (long long i = 0; i < n; i++) {
    scanf("%lld", &input[i]);
  }
  if (n == 0) {
    printf("0");
    return 0;
  }
  map<int, long long> m;
  m[4] = 0;
  m[8] = 0;
  m[15] = 0;
  m[16] = 0;
  m[23] = 0;
  m[42] = 0;
  for (long long i = 0; i < n; i++) {
    for (auto subIt = m.begin(); subIt != m.end(); ++subIt) {
      if (input[i] == subIt->first) {
        long long minCount = 9223372036854775807;
        for (auto subIt2 = m.begin(); subIt2 != subIt; ++subIt2) {
          minCount = (minCount < subIt2->second) ? minCount : subIt2->second;
        }
        if (subIt->second + 1 <= minCount) {
          subIt->second++;
        }
      }
    }
  }
  long long numSub = m[42];
  long long extra = 0;
  long long marked = 0;
  for (auto it = m.begin(); it != m.end(); ++it) {
    marked += it->second;
    extra += (it->second - numSub);
  }
  long long unmarked = n - marked;
  printf("%lld", extra + unmarked);
  return 0;
}
