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
import heapq as h

n = int(input())
arr = list(map(int, input().split()))

vals = [4, 8, 15, 16, 23, 42]

list = {}
for val in vals:
    list[val] = []

for idx in range(n):
    list[arr[idx]].append(idx)

for val in vals:
    h.heapify(list[val])

used = [False] * n

while len(list[4]) != 0:
    seq = [list[4][0]]
    h.heappop(list[4])

    for idx in range(1, len(vals)):
        while len(list[vals[idx]]) != 0 and list[vals[idx]][0] < seq[-1] and not used[list[vals[idx]][0]]:
            h.heappop(list[vals[idx]])

        if len(list[vals[idx]]) != 0:
            seq.append(list[vals[idx]][0])
            h.heappop(list[vals[idx]])
        else:
            break

    if len(seq) < 6:
        break

    for no in seq:
        used[no] = True

cnt = 0
for no in used:
    if no == False:
        cnt += 1

print(cnt)