Problem:
Score : 400 points

Problem Statement

You start with the number 0 and you want to reach the number N.

You can change the number, paying a certain amount of coins, with the following operations:

Multiply the number by 2, paying A coins.

Multiply the number by 3, paying B coins.

Multiply the number by 5, paying C coins.

Increase or decrease the number by 1, paying D coins.

You can perform these operations in arbitrary order and an arbitrary number of times.

What is the minimum number of coins you need to reach N?

You have to solve T testcases.

Constraints

1 \le T \le 10

1 \le N \le 10^{18}

1 \le A, B, C, D \le 10^9

All numbers N, A, B, C, D are integers.

Input

The input is given from Standard Input. The first line of the input is

T

Then, T lines follow describing the T testcases.
Each of the T lines has the format

N A B C D

Output

For each testcase, print the answer on Standard Output followed by a newline.

Sample Input 1

5
11 1 2 4 8
11 1 2 2 8
32 10 8 5 4
29384293847243 454353412 332423423 934923490 1
900000000000000000 332423423 454353412 934923490 987654321

Sample Output 1

20
19
26
3821859835
23441258666

For the first testcase, a sequence of moves that achieves the minimum cost of 20 is:

Initially x = 0.

Pay 8 to increase by 1 (x = 1).

Pay 1 to multiply by 2 (x = 2).

Pay 1 to multiply by 2 (x = 4).

Pay 2 to multiply by 3 (x = 12).

Pay 8 to decrease by 1 (x = 11).

For the second testcase, a sequence of moves that achieves the minimum cost of 19 is:

Initially x = 0.

Pay 8 to increase by 1 (x = 1).

Pay 1 to multiply by 2 (x = 2).

Pay 2 to multiply by 5 (x = 10).

Pay 8 to increase by 1 (x = 11).

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <cstdio>
#include <map>
using namespace std;
typedef long long ll;

ll T, n, a, b, c, d;
map<ll, ll> mp;

ll f(ll p) {
	if(mp.count(p)) return mp[p];
	ll re = 1e18;
	if(p*d>0) re = min(re, p*d);
	re = min(re, f(p/2) + a + (p%2)*d);
	re = min(re, f(p/3) + b + (p%3)*d);
	re = min(re, f(p/5) + c + (p%5)*d);
	if(p>2)	re = min(re, f(p/2+1) + a + (2-p%2)*d);
	re = min(re, f(p/3+1) + b + (3-p%3)*d);
	re = min(re, f(p/5+1) + c + (5-p%5)*d);
	return mp[p] = re;
}

int main() {
	cin >> T;
	while(T--) {
		cin >> n >> a >> b >> c >> d;
		mp.clear();
		mp[0] = 0;
		mp[1] = d;
		cout << f(n) << endl;
	}
	return 0;
}