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)

# Inputing values


t=int(input())

# for loop 

for i in range(t):
    n,a,b,c,d=map(int,input().split())
n,A,B,C=map(int,input().split())

# new vars


l=[int(input()) for _ in range(n)]


# defineing new 


def dfs(count,a,b,c):
  
  # if-else looping 
    if count==n:
        if min(a,b,c)>0:
            return abs(a-A)+abs(b-B)+abs(c-C)-30
        else:
            return 10**9+1
          
          # new vars
          
    x = dfs(count+1, a, b, c)
    y = dfs(count+1, a+l[count], b, c)+10
    z = dfs(count+1, a, b+l[count], c)+10
    w = dfs(count+1, a, b, c+l[count])+10
    return min(x,y,z,w)

# printing answer 

print(dfs(0,0,0,0))