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
import sys
input = sys.stdin.readline
sys.setrecursionlimit(2147483647)
def main():
    T = int(input())
    for _ in range(T):
        N, A, B, C, D = map(int, input().split())
        memo = {}

        def f(n):
            if n == 0:return 0
            if n == 1:return D
            if n in memo:return memo[n]

            ret = min(
                D * n,
                D * (n%2) + A + f(n//2),
                D * abs(n - (n+1)//2*2) + A + f((n+1)//2),
                D * (n%3) + B + f(n//3),
                D * abs(n - (n+2)//3*3) + B + f((n+2)//3),
                D * (n%5) + C + f(n//5),
                D * abs(n - (n+4)//5*5) + C + f((n+4)//5)
            )

            memo[n] = ret
            return ret
        print(f(N))
if __name__ == "__main__":
    main()