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 Python (3.8.2) (judge verdict: Accepted)

from functools import lru_cache

T = int(input())


def resolver(a, b, c, d):
    @lru_cache(maxsize=None)
    def _resolver(n):
        if n == 0:
            return 0
        elif n == 1:
            return d
        else:
            div5f = n // 5
            div5c = (n + 4) // 5
            div3f = n // 3
            div3c = (n + 2) // 3
            div2f = n // 2
            div2c = (n + 1) // 2
            return min(
                d * n,
                d * abs(n - div5f * 5) + c + _resolver(div5f),
                d * abs(n - div5c * 5) + c + _resolver(div5c),
                d * abs(n - div3f * 3) + b + _resolver(div3f),
                d * abs(n - div3c * 3) + b + _resolver(div3c),
                d * abs(n - div2f * 2) + a + _resolver(div2f),
                d * abs(n - div2c * 2) + a + _resolver(div2c),
            )

    return _resolver


for _ in range(T):
    N, A, B, C, D = map(int, input().split(' '))
    print(resolver(A, B, C, D)(N))