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: Wrong Answer)

import heapq
class DFS:
    def __init__(self,n,a,b,c,d):
        self.N = n
        self.A = a
        self.B = b
        self.C = c
        self.D = d
        self.ans = self.N * self.D
        self.Q = [[0,n]]
    def dfs(self):
        while self.Q:
#             from IPython.core.debugger import Pdb; Pdb().set_trace()
            cost,num = heapq.heappop(self.Q)
            if num == 0:
                return cost
            elif num == 1:
                heapq.heappush(self.Q,[cost+self.D,0])
            else:
                # 5の倍数に上げて5で割った場合
                temp_num = num//5 + 1
                temp_cost = cost + ((num//5 + 1) * 5 - num) * self.D + self.C
                heapq.heappush(self.Q,[temp_cost,temp_num])

                # 5の倍数に下げて5で割った場合
                temp_num = num//5
                temp_cost = cost + (num - (num//5)*5) * self.D + self.C
                heapq.heappush(self.Q,[temp_cost,temp_num])

                # 3の倍数に上げて3で割った場合
                temp_num = num//3 + 1
                temp_cost = cost + ((num//3 + 1) * 3 - num) * self.D + self.B
                heapq.heappush(self.Q,[temp_cost,temp_num])

                # 3の倍数に下げて3で割った場合
                temp_num = num//3
                temp_cost = cost + (num - (num//3)*3) * self.D + self.B
                heapq.heappush(self.Q,[temp_cost,temp_num])

                # 2の倍数に上げて2で割った場合
                temp_num = num//2 + 1
                temp_cost = cost + ((num//2 + 1) * 2 - num) * self.D + self.A
                heapq.heappush(self.Q,[temp_cost,temp_num])

                # 2の倍数に下げて2で割った場合
                temp_num = num//2
                temp_cost = cost + (num - (num//2)*2) * self.D + self.A
                heapq.heappush(self.Q,[temp_cost,temp_num])

                # 全て引いた場合
                heapq.heappush(self.Q,[temp_cost,temp_num])
                
        return self.ans
T = int(input())
for _ in range(T):
    N,A,B,C,D = map(int,input().split())
    case = DFS(N,A,B,C,D)
    print(case.dfs())