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
from collections import defaultdict as dd
import heapq
hpush = heapq.heappush
hpop = heapq.heappop
input = sys.stdin.readline
for _ in range(int(input())):
  N, A, B, C, D = map(int, input().split())
  h = [(0, N)]
  d = dd(lambda: float("inf"))
  d[N] = 0
  vis = set()
  while len(h):
    c, x = hpop(h)
    if x in vis: continue
    vis.add(x)
    if x == 0:
      print(c)
      break
    #print(h)
    if d[0] <= c: continue
    hpush(h, (c + x * D, 0))
    d[0] = min(d[0], c + x * D)
    if d[x // 2] > c + A + (x % 2) * D:
      hpush(h, (c + A + (x % 2) * D, x // 2))
      d[x // 2] = c + A + (x % 2) * D
    if d[-(-x // 2)] > c + A + (x % 2) * D:
      hpush(h, (c + A + (-x % 2) * D, -(-x // 2)))
      d[-(-x // 2)] = c + A + (-x % 2) * D

    if d[x // 3] > c + B + (x % 3) * D:
      hpush(h, (c + B + (x % 3) * D, x // 3))
      d[x // 3] = c + B + (x % 3) * D
    if d[-(x // 3)] > c + B + (-x % 3) * D:
      hpush(h, (c + B + (-x % 3) * D, -(-x // 3)))
      d[-(-x // 3)] = c + B + (-x % 3) * D

    if d[x // 5] > c + C + (x % 5) * D:
      hpush(h, (c + C + (x % 5) * D, x // 5))
      d[x // 5] = c + C + (x % 5) * D
    if d[x // 5] > c + C + (-x % 5) * D:
      hpush(h, (c + C + (-x % 5) * D, -(-x // 5)))
      d[-(-x // 5)] = c + C + (-x % 5) * D