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;
using lli = long long;
void fast() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
long long ncr(int n, int r) {
  if (r > n - r) r = n - r;
  long long ans = 1;
  lli j = 2;
  for (int i = n - r + 1; i <= n; i++) {
    ans *= i;
    while (!(ans % j) && j <= r) {
      ans /= j;
      j++;
    }
  }
  return ans;
}
lli gcd(lli a, lli b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
lli lcm(lli n, lli m) { return n * m / gcd(n, m); }
bool isprime(lli n) {
  if (n == 0 || n == 1) return 0;
  if (n == 2 || n == 3) return 1;
  if (n % 2 == 0 || n % 3 == 0) return 0;
  for (int i = 5; i <= sqrt(n); i += 6)
    if (n % i == 0 || n % (i + 2) == 0) return 0;
  return 1;
}
bool vis[2000001];
lli len, sp[2000001];
lli cnt, sum, mn = 1e9, ans, mxx = 0, mx;
vector<lli> primes;
lli frq[3];
lli n, mem[2000001][2];
vector<lli> a(2000001), b(2000001);
int main() {
  fast();
  lli q;
  q = 1;
  while (q--) {
    map<lli, lli> mp;
    lli n, x;
    cin >> n;
    vector<lli> good = {4, 8, 15, 16, 23, 42};
    for (int i = 0; i < n; i++) {
      cin >> x;
      auto z = lower_bound(good.begin(), good.end(), x) - good.begin();
      if (z == 0 || mp[x] < mp[good[z - 1]]) mp[x]++;
    }
    cout << n - mp[42] * 6 << "\n";
  }
}
