Codeforces Round #565 (Div. 3)C. Lose it!

You are given an array a consisting of n integers. Each ai 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 k6 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⋅105) — the number of elements in a.

The second line of the input contains n integers a1,a2,…,an (each ai is one of the following numbers: 4,8,15,16,23,42), where ai 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
inputCopy
5
4 8 15 16 23
outputCopy
5
inputCopy
12
4 8 4 15 16 8 23 15 16 42 23 42
outputCopy
0
inputCopy
15
4 8 4 8 15 16 8 16 23 15 16 4 42 23 42
outputCopy
3

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<ctime>
#include<iostream>
#include<algorithm>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<string>
#define ll long long
#define dd double
using namespace std;

int main() {
	ios::sync_with_stdio(0);
	ll t;
	cin >> t;
	while (t--) {
		ll n; cin >> n;
		ll sum = 0;
		ll flag = 0;
		while (n != 1) {
			if (n % 2 == 0) {
				n = n / 2;
				sum++;
			}
			else if(n % 3 == 0 && n % 5 == 0) {
				n = min(((n / 3) * 2),((n / 5) * 4));
				sum++;
			}
			else if (n % 3 == 0) {
				n = n / 3 * 2;
				sum++;
			}
			else if (n % 5 == 0) {
				n = n / 5 * 4;
				sum++;
			}
			else {
				flag = 1;
				cout << "-1" << endl;
				break;
			}
		}
		if (flag == 0) {
			cout << sum << endl;
		}
	}
}