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 PyPy3 (7.3.0) (judge verdict: Accepted)

import sys
input = sys.stdin.readline
from functools import lru_cache

T=int(input())
for tests in range(T):
    N,A,B,C,D=map(int,input().split())

    DP=[N*D]*(10**5)
    DP[0]=0

    for i in range(10**5):
        for k in range(i-10,i+10):
            if 0<=k<10**5 and DP[k]>DP[i]+abs(i-k)*D:
                DP[k]=DP[i]+abs(i-k)*D
        #if 0<=i+1<10**5 and DP[i+1]>DP[i]+D:
        #    DP[i+1]=DP[i]+D
            
        #if 0<=i-1<10**5 and DP[i-1]>DP[i]+D:
        #    DP[i-1]=DP[i]+D

        if 0<=2*i<10**5 and DP[2*i]>DP[i]+A:
            DP[2*i]=DP[i]+A

        if 0<=3*i<10**5 and DP[3*i]>DP[i]+B:
            DP[3*i]=DP[i]+B

        if 0<=5*i<10**5 and DP[5*i]>DP[i]+C:
            DP[5*i]=DP[i]+C

    @lru_cache(maxsize=None)
    def calc(x):
        if x<=10:
            return DP[x]

        ANS=x*D

        two=x//2*2

        ANS=min(ANS,abs(two-x)*D+A+calc(two//2))
        ANS=min(ANS,abs(two+2-x)*D+A+calc(two//2+1))

        three=x//3*3
        ANS=min(ANS,abs(three-x)*D+B+calc(three//3))
        ANS=min(ANS,abs(three+3-x)*D+B+calc(three//3+1))

        five=x//5*5
        ANS=min(ANS,abs(five-x)*D+C+calc(five//5))
        ANS=min(ANS,abs(five+5-x)*D+C+calc(five//5+1))

        return ANS

    print(calc(N))