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 Ruby (2.7.1) (judge verdict: Accepted)

class Problem
  MOD = 10007
  MAX = 10 ** 12
  attr_reader :a, :b, :c, :d, :memo, :cost

  def initialize(a, b, c, d)
    @a = a; @b = b; @c = c; @d = d
    @memo = { 0 => 0, 1 => @d }
    @cost = { 2 => @a, 3 => @b, 5 => @c}
  end

  def solve(n)
    return memo[n] if memo[n]
    mini = MAX
    candidates(n) do |to, cost|
      now = solve(to) + cost
      mini = now if now < mini
    end
    memo[n] = mini
  end

  def candidates(n)
    yield [0, d * n]
    [2, 3, 5].each do |i|
      j = cost[i]
      if i <= n
        yield [n / i, j + d * (n % i)]
      end

      if n % i != 0
        yield [(n + i - 1) / i, j + d * (i - n % i)]
      end
    end
  end
end

t = gets.to_i
t.times do
  n, a, b, c, d = gets.split.map(&:to_i)
  pr = Problem.new(a, b, c, d)
  puts pr.solve(n)
end