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.

Example

Input

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


Output

20
19
26
3821859835
23441258666
from functools import lru_cache
T = int(input())
for _ in range(T):
    @lru_cache(maxsize=None)
    def f(n):
        if n == 0:
            return 0
        if n == 1:
            return D
        if n == 2:
            return min(2*D, A+D)
        temp = [D * n]
        d = n % 2
        temp.append(f(n // 2) + A + D * d)
        temp.append(f(n // 2 + 1) + A + D * (2-d))
        d = n % 3
        temp.append(f(n // 3) + B + D * d)
        temp.append(f(n // 3 + 1) + B + D * (3-d))
        d = n % 5
        temp.append(f(n // 5) + C + D * d)
        temp.append(f(n // 5 + 1) + C + D * (5-d))
        return min(temp)
    N, A, B, C, D = map(int, input().split())
    print(f(N))
