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 Java (OpenJDK 11.0.6) (judge verdict: Runtime Error)

package AGC44;
import java.util.Scanner;

class Main {
	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);

		int T = Integer.parseInt(sc.next());

		long[][] ca = new long[T][5];
		long[] ans = new long[T];
		for(int i = 0;i < T;i++) {
			 ca[i][0] = Long.parseLong(sc.next());
			 ca[i][1] = Long.parseLong(sc.next());
			 ca[i][2] = Long.parseLong(sc.next());
			 ca[i][3] = Long.parseLong(sc.next());
			 ca[i][4] = Long.parseLong(sc.next());

			 ans[i] = calc(ca[i]);
		}

		for(long a : ans) {
			System.out.println(a);
		}
	}

	public static long calc(long[] a) {
			 long total = 0;
			 long zan = 0;

			 long[] bb = new long[5];

			 while(zan != a[0]) {
				 if(zan == 0) {
					 zan++;
					 total += a[4];
				 }

				 bb[0] = Math.abs(a[0]-(zan * 2));
				 bb[1] = Math.abs(a[0]-(zan * 3));
				 bb[2] = Math.abs(a[0]-(zan * 5));
				 bb[3] = Math.abs(a[0]-(++zan ));
				 bb[4] = Math.abs(a[0]-(--zan ));

				 long min =Long.MAX_VALUE;
				 int ii = 0;
				 for(int i = 0; i < bb.length;i++) {
					 if(min >= bb[i]) {
						 min = bb[i];
						ii = i;
					 }
				 }
				 switch(ii) {
				 case 0:
					 zan = zan * 2;
					 total += a[1];
					 break;
				 case 1:
					 zan = zan * 3;
					 total += a[2];
					 break;
				 case 2:
					 zan = zan * 5;
					 total += a[3];
					 break;
				 case 3:
					 zan++;
					 total += a[4];
					 break;
				 case 4:
					 zan--;
					 total += a[4];
					 break;
				 }

			 }
		return total;

	}

}