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: Time Limit Exceeded)

'''
T = 5
N = 11
A = 1#*2
B = 2#*3
C = 4#*5
D = 8#+1
'''


T = int(input())
#T = 1
for _ in range(T):
	N, A, B, C, D = map(int, input().split())
	#[A, B, C, D]
	memo = [0, D]#{1: D}
	for i in range(2, N+2):
		cost = memo[i-1] + D
		if i%2==0:
			cost = min(cost, memo[int(i/2)] + A)
		if i%3==0:
			cost = min(cost, memo[int(i/3)] + B)
		if i%5==0:
			cost = min(cost, memo[int(i/5)] + C)
		#memo[i] = cost
		memo.append(cost)
		memo[i-1] = min(memo[i-1], cost + D)
		#print(str(i)+' >>> '+str(cost))
	print(memo[N])
print()