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
#Bhargey Mehta (Sophomore)
#DA-IICT, Gandhinagar
import sys, math, queue, bisect
#sys.stdin = open("input.txt", "r")
MOD = 10**9+7
sys.setrecursionlimit(1000000)

n = int(input())
a = list(map(int, input().split()))
d = {4: [], 8: [], 15: [], 16: [], 23:[], 42:[]}
s = [4, 8, 15, 16, 23, 42]
for i in range(n):
    d[a[i]].append(i)
for k in d:
    d[k].reverse()
ans = 0
while True:
    last = None
    pop = 0
    for si in s:
        if last == None:
            if len(d[si]) == 0:
                for k in d:
                    ans += len(d[k])
                    d[k] = []
                break
            last = d[si].pop()
            pop += 1
        elif len(d[si]) == 0:
            ans += pop
            break
        else:
            while len(d[si]) > 0 and last > d[si][-1]:
                ans += 1
                d[si].pop()
            if len(d[si]) == 0:
                ans += pop
                break
            else:
                last = d[si].pop()
                pop += 1
    over = True
    for k in d:
        if len(d[k]) != 0:
            over = False
    if over: break
print(ans)
