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: Time Limit Exceeded)

import sys
import math

ceil = math.ceil

sys.setrecursionlimit(10**6)
readline = sys.stdin.readline
#read = sys.stdin.buffer.read

def f(n, div, cnt):
    global A, B, C, D, dct
    fltmp = n//div
    cltmp = ceil(n/div)
    if fltmp in [0, 1] and cltmp == 1:
        cntfl = cnt + n*D
        cntcl = cnt + abs(n-div*cltmp)*D + (div==2)*A + (div==3)*B + (div==5)*C + D
        if 0 not in dct or dct[0] > cntfl or dct[0] > cntcl:
            dct[0] = min(cntfl, cntcl)
        return
    cntfl = cnt + abs(n-div*fltmp)*D + (div==2)*A + (div==3)*B + (div==5)*C
    cntcl = cnt + abs(n-div*cltmp)*D + (div==2)*A + (div==3)*B + (div==5)*C
    cnt0fl = cntfl + fltmp*D
    cnt0cl = cntcl + cltmp*D
    if 0 not in dct or dct[0] > cnt0fl or dct[0] > cnt0cl:
        dct[0] = min(cnt0cl, cnt0fl)
    if fltmp not in dct or dct[fltmp] > cntfl:
        dct[fltmp] = cntfl
        for divtmp in [2, 3, 5]:
            f(fltmp, divtmp, cntfl)
    if cltmp not in dct or dct[cltmp] > cntcl:
        dct[cltmp] = cntcl
        for divtmp in [2, 3, 5]:
            f(cltmp, divtmp, cntcl)

def main():
    global A, B, C, D, dct
    T = int(input())
    for _ in range(T):
        N, A, B, C, D = map(int, readline().split())
        dct = {N:0}
        for divtmp in [2, 3, 5]:
            f(N, divtmp, 0)
        ans = min(N*D, dct[0])
        print(ans)


if __name__ == "__main__":
    main()