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: Wrong Answer)


import java.util.*;

public class Main {
    Scanner in = new Scanner(System.in);

    void handle(long[] curr, long m, long cost, long d, PriorityQueue<long[]> q, Map<Long, Long> visited) {
        long c = curr[0] + cost + d * (curr[1] % m);
        if (!visited.containsKey(curr[1]/m) || visited.get(curr[1]/m) > c) {
            q.add(new long[]{c, curr[1]/m});
            visited.put(curr[1]/m, c);
        }

        c = curr[0] + cost + d * (m-curr[1] % m);
        if (!visited.containsKey(curr[1] / m + 1) || visited.get(curr[1] / m + 1) > c) {
            q.add(new long[]{c, curr[1] / m + 1});
            visited.put(curr[1] / m + 1, c);
        }
    }

    long solve() {
        long n = in.nextLong(), a = in.nextLong(), b = in.nextLong(), c = in.nextLong(), d = in.nextLong();
        PriorityQueue<long[]> q = new PriorityQueue<>(Comparator.comparingLong(x -> x[0]));
        q.add(new long[]{0, n});
        long ret = -1;
        Map<Long, Long> visited = new HashMap<>();
        while (!q.isEmpty()) {
            long[] curr = q.poll();
            if (curr[1] == 0) {
                ret = curr[0];
                break;
            }

            handle(curr, 2, a, d, q, visited);
            handle(curr, 3, b, d, q, visited);
            handle(curr, 5, c, d, q, visited);

            if (curr[1] == 1) {
                if (!visited.containsKey(0) || visited.get(0) < curr[0] + d) {
                    visited.put(0L, curr[0] + d);
                    q.add(new long[]{curr[0] + d, 0});
                }
            }
        }

        return Math.min(ret, d * n);
    }

    void run() {
        int T = in.nextInt();
        while (T-- > 0) {
            System.out.println(solve());
        }
    }

    public static void main(String[] args) {
        new Main().run();
    }
}