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: Wrong Answer)

import sys
sys.setrecursionlimit(10 ** 9)
t = int(input())
for _ in range(t):
    n, a, b, c, d = map(int, input().split())
    memo = {i: -1 for i in range(n+1)}
    def f(x):
        if memo[x] != -1:
            return memo[x]
        if x == 0:
            return 0
        if abs(x) == 1:
            return d

        ans = []
        if x % 2 == 0:
            ans.append(f(x // 2) + a)
        else:
            ans.append(f((x+1) // 2) + a + d)
            ans.append(f((x-1) // 2) + a + d)

        if x % 3 == 0:
            ans.append(f(x // 3) + b)
        elif x % 3 == 1:
            ans.append(f((x-1) // 3) + b + d)
        elif x % 3 == 2:
            ans.append(f((x+1) // 3) + b + d)

        if x % 5 == 0:
            ans.append(f(x // 5) + c)
        elif x % 5 == 1:
            ans.append(f((x-1) // 5) + c + d)
        elif x % 5 == 2:
            ans.append(f((x-2) // 5) + c + 2 * d)
        elif x % 5 == 3:
            ans.append(f((x+2) // 5) + c + 2 * d)
        elif x % 5 == 4:
            ans.append(f((x+1) // 5) + c + d)

        memo[x] = min(ans)
        return memo[x]

    print(f(n))