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: Accepted)

import sys
from heapq import heappush, heappop
T = int(input())
for _ in range(T):
	n, a, b, c, d = map(int, input().split())
	dp = dict()
	dp[n] = 0
	q = [(0, n)]
	def add(x, v):
		if x not in dp:
			dp[x] = v
			return True
		elif dp[x] > v:
			dp[x] = v
			return True
		else:
			return False

	while True:
		l = heappop(q)
		if l[1] == 0:
			break
		x = l[1]
		if x%2 == 0:
			v = min(a, d*(x//2)) + dp[x]
			if add(x//2, v):
				heappush(q, (v, x//2))
		else:
			v = min(a+d, d + d*(x//2)) + dp[x]
			if add(x//2, v):
				heappush(q, (v, x//2))
			v = min(a+d, d + d*(x//2+1)) + dp[x]
			if add(x//2+1, v):
				heappush(q, (v, x//2+1))

		if x%3 == 0:
			v = min(b, d*2*(x//3)) + dp[x]
			if add(x//3, v):
				heappush(q, (v, x//3))
		else:
			m = x%3
			v = min(b+d*m, d*m + d*2*(x//3)) + dp[x]
			if add(x//3, v):
				heappush(q, (v, x//3))
			v = min(b+d*(3-m), d*(3-m) + d*2*(x//3 + 1)) + dp[x]
			if add(x//3+1, v):
				heappush(q, (v, x//3+1))

		if x%5 == 0:
			v = min(c, d*4*(x//5)) + dp[x]
			if add(x//5, v):
				heappush(q, (v, x//5))
		else:
			m = x%5
			v = min(c+d*m, d*m + d*4*(x//5)) + dp[x]
			if add(x//5, v):
				heappush(q, (v, x//5))
			v = min(c+d*(5-m), d*(5-m) + d*4*(x//5 + 1)) + dp[x]
			if add(x//5+1, v):
				heappush(q, (v, x//5+1))

	print(dp[0])