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: Runtime Error)

from math import ceil,floor,log
from decimal import *
from heapq import *
import sys
input=sys.stdin.readline
T=int(input())
cases=[list(map(int,input().split())) for _ in range(T)]
getcontext().prec=20
num=0
cnt=0
for i in cases:
    l2=[0]*3
    N,l2[0],l2[1],l2[2],D=i
    d = {}
    num=0
    cnt=0
    for i in range(ceil(log(N,2))):         #2の累乗
        for j in range(ceil(log(N,3))):     #3の累乗
            for k in range(ceil(log(N,5))): #5の累乗
                div=pow(2,i)*pow(3,j)*pow(5,k)
                x=Decimal(N)/Decimal(div)
                c=x.to_integral_exact(rounding=ROUND_CEILING)
                f=x.to_integral_exact(rounding=ROUND_FLOOR)
                if d.get(c)==None:
                    d[c]=float("inf")
                if d.get(f)==None:
                    d[f]=float("inf")
    d[N]=0
    #print(d)
    q=[]
    q.append((0,N))
    heapify(q)
    while len(q)>0:
        dist,u=heappop(q)
        if d[u]<dist: continue
        for j in range(3):
            num=i+pow(2,j)
            cnt=l2[j]
            x = Decimal(u) / Decimal(num)
            v1 = x.to_integral_exact(rounding=ROUND_CEILING)
            v2 = x.to_integral_exact(rounding=ROUND_FLOOR)
            # v1=ceil(u/2)
            # v2=floor(u/2)
            alt1 = d[u] + min(cnt+((num-(u%num))%num)*D,(u-v1)*D)
            alt2 = d[u] + min(cnt+(u%num)*D,(u-v2)*D)
            if d[v1] > alt1:
                d[v1] = alt1
                heappush(q, (alt1, v1))
            if d[v2] > alt2:
                d[v2] = alt2
                heappush(q, (alt2, v2))
        #print(d)
        #input()
    print(d[0])
    #print(len(d))