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: Accepted)

import java.util.*;

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

    long f(long n, long[] costs, long d, Map<Long, Long> memo) {
        if (n == 0) return 0;
        if (n == 1) return d;
        if (memo.containsKey(n)) return memo.get(n);

        int[] div = new int[]{2,3,5};
        long ret = Long.MAX_VALUE;
        if (ret / d >= n) ret = n * d;
        for (int i=0;i<3;i++) {
            int di = div[i];
            long c = costs[i];
            if (n % di == 0) {
                ret = Math.min(ret, f(n/di, costs, d, memo) + c);
            } else {
                ret = Math.min(ret, c + f(n / di, costs, d, memo) + d * (n % di));
                ret = Math.min(ret, c + f(n / di + 1, costs, d, memo) + d * (di - n % di));
            }
        }

        memo.put(n, ret);
        return ret;
    }

    long solve() {
        long n = in.nextLong(), a = in.nextLong(), b = in.nextLong(), c = in.nextLong(), d = in.nextLong();
        return f(n, new long[]{a, b, c}, d, new HashMap<>());
    }

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

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