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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

t = int(input())
import sys
sys.setrecursionlimit(10**6)

seen = set()
dp = dict()
def Find(value, isFirst = False):
    seen.add(value)

    if value == 1:
        return d
    elif value == 2:
        return min(Find(1) + d, Find(1) + a)
    elif value == 3:
        return min(Find(2) + d, Find(1) + b)
    elif value == 4:
        return min(Find(3) + d, Find(2) + a)
    elif value == 5:
        return min(Find(4) + d, Find(1) + c)
    elif value == 6:
        return min(Find(5) + d, Find(2) + b, Find(3) + a)
    else:
        if value in dp:
            return dp[value]

        candidate = []

        if isFirst:
            candidate.append(Find(value + 1) + d)
        if value % 2 == 0:
            candidate.append(Find(value//2) + a)
        if value % 3 == 0:
            candidate.append(Find(value//3) + b)
        if value % 5 == 0:
            candidate.append(Find(value//5) + c)
        if value % 6 == 0:
            candidate.append(Find(value // 2) + a)
            candidate.append(Find(value // 3) + b)
            candidate.append((value-2)*d)
        else:
            candidate.append(Find(value - 1) + d)


        if len(candidate) == 0:
            raise 1
        min1 = min(candidate)
        dp[value] = min1
        return min1

for _ in range(t):
    n, a, b, c, d =map(int, input().split())
    seen.clear()
    dp.clear()
    v = Find(n, True)
    print(v)