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)

def solve(N, A, B, C, D):
    d = {0:0, 1:D}
    s = set()
    return dfs(N, A, B, C, D, d, s)

def dfs(n, A, B, C, D, d, s):
    if n in d:
        return d[n]
    if n in s:
        return float('inf')
    
    s.add(n)
    ans = float('inf')
    
    p = n % 2
    if p == 0:
        ans = min(ans, dfs(n // 2, A, B, C, D, d, s) + A)
    else:
        ans = min(ans, dfs((n - p) // 2, A, B, C, D, d, s) + A + D * p)
        ans = min(ans, dfs((n + (2 - p)) // 2, A, B, C, D, d, s) + A + D * (2 - p))

    q = n % 3
    if q == 0:
        ans = min(ans, dfs(n // 3, A, B, C, D, d, s) + B)
    else:
        ans = min(ans, dfs((n - q) // 3, A, B, C, D, d, s) + B + D * q)
        ans = min(ans, dfs((n + (3 - q)) // 3, A, B, C, D, d, s) + B + D * (3 - q))
    
    r = n % 5
    if r == 0:
        ans = min(ans, dfs(n // 5, A, B, C, D, d, s) + C)
    else:
        ans = min(ans, dfs((n - r) // 5, A, B, C, D, d, s) + C + D * r)
        ans = min(ans, dfs((n + (5 - r)) // 5, A, B, C, D, d, s) + C + D * (5 - r))
    
    ans = min(ans, D * n)
    
    d[n] = ans
    return ans

T = int(input())
for _ in range(T):
    N, A, B, C, D = map(int,input().split())
    ans = solve(N, A, B, C, D)
    print(ans)