output_description: For each testcase, print the answer on Standard Output followed by a newline.
  
  * * *
input_description: 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
attempt: import sys
  from collections import deque
  from typing import Callable, NoReturn
  
  
  def bfs(n: int, a: int, b: int, c: int, d: int) -> int:
      queue = deque([(1, d)])
  
      while queue:
          num, coins = queue.popleft()  # type: int, int
  
          if n == num:
              return coins
  
          queue.append((num * 2, coins + a))
          queue.append((num * 3, coins + b))
          queue.append((num * 5, coins + c))
  
  
  def main() -> NoReturn:
      readline: Callable[[], str] = sys.stdin.readline
  
      t = int(readline().rstrip())
  
      for _ in range(t):
          n, a, b, c, d = map(int, readline().rstrip().split())  # type: int,...
          print(bfs(n, a, b, c, d))
  
  
  if __name__ == "__main__":
      main()
  
problem_description: 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.**
samples: [{"input": "5\n    11 1 2 4 8\n    11 1 2 2 8\n    32 10 8 5 4\n    29384293847243 454353412 332423423 934923490 1\n    900000000000000000 332423423 454353412 934923490 987654321", "output": "20\n    19\n    26\n    3821859835\n    23441258666\n    \n\nFor the first testcase, a sequence of moves that achieves the minimum cost of\n20 is:\n\n  * Initially x = 0.\n  * Pay 8 to increase by 1 (x = 1).\n  * Pay 1 to multiply by 2 (x = 2).\n  * Pay 1 to multiply by 2 (x = 4).\n  * Pay 2 to multiply by 3 (x = 12).\n  * Pay 8 to decrease by 1 (x = 11).\n\nFor the second testcase, a sequence of moves that achieves the minimum cost of\n19 is:\n\n  * Initially x = 0.\n  * Pay 8 to increase by 1 (x = 1).\n  * Pay 1 to multiply by 2 (x = 2).\n  * Pay 2 to multiply by 5 (x = 10).\n  * Pay 8 to increase by 1 (x = 11)."}]