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;
  cin >> n;
  long long* a = new long long[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  if (n < 6) {
    cout << n;
    return 0;
  }
  long long dp[6][n];
  long long* freq = new long long[6];
  for (int i = 0; i < 6; i++) freq[i] = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == 4) {
      dp[0][freq[0]] = i;
      freq[0]++;
    }
    if (a[i] == 8) {
      dp[1][freq[1]] = i;
      freq[1]++;
    }
    if (a[i] == 15) {
      dp[2][freq[2]] = i;
      freq[2]++;
    }
    if (a[i] == 16) {
      dp[3][freq[3]] = i;
      freq[3]++;
    }
    if (a[i] == 23) {
      dp[4][freq[4]] = i;
      freq[4]++;
    }
    if (a[i] == 42) {
      dp[5][freq[5]] = i;
      freq[5]++;
    }
  }
  int i = 0, j = 0, k = 0, l = 0, m = 0, n1 = 0;
  long long count = 0;
  while (i < freq[0] && j < freq[1] && k < freq[2] && l < freq[3] &&
         m < freq[4] && n1 < freq[5]) {
    long long start = dp[0][i];
    i++;
    while (dp[1][j] < start && j < freq[1]) {
      j++;
    }
    if (j >= freq[1]) break;
    start = dp[1][j];
    j++;
    while (dp[2][k] < start && k < freq[2]) {
      k++;
    }
    if (k >= freq[2]) break;
    start = dp[2][k];
    k++;
    while (dp[3][l] < start && l < freq[3]) {
      l++;
    }
    if (l >= freq[3]) break;
    start = dp[3][l];
    l++;
    while (dp[4][m] < start && m < freq[4]) {
      m++;
    }
    if (m >= freq[4]) break;
    start = dp[4][m];
    m++;
    while (dp[5][n1] < start && n1 < freq[5]) {
      n1++;
    }
    if (n1 >= freq[5]) break;
    n1++;
    count++;
  }
  long long ans = n - (6 * count);
  cout << ans << endl;
}
