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: Wrong Answer)

#include  <iostream>

int main(void)
{
	int T;
	int n, a, b, c, d;
	int var = 0;

	std::cin >> T;

	for(int i = 0; i < T; i++){
		int pay = 0;
		int unit_a, unit_b, unit_c;
		std::cin >> n >> a >> b >> c >> d;

		unit_a = a/2;
		unit_b = b/3;
		unit_c = c/5;

		int maxpay, maxunit;
		int minpay, minunit;

		int amp;

		pay += d;
		var = 1;

		while(1){
			if(var == n){
				std::cout << pay << std::endl;
				break;
			}

			amp = n / var;

			if(var < n){
			    if( amp == 1 ){
			    	pay += (n - var) * d;
			    	var += (n - var);
			    }
			    else if(amp == 2){
			    	pay += a;
			    	var *= 2;
			    }
			    else if(amp == 3){
			    	pay += b;
			    	var *= 3;
			    }
			    else if(amp == 5){
			    	pay += c;
			    	var *= 5;
			    }
			}
			else{
				pay += (var - n)*d;
				var -= (var - n);
			}
		}

	}

	return 0;
}