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: Wrong Answer)

class Entry
  attr_reader :n, :a, :b, :c, :d, :cache

  def initialize(line)
    @n, @a, @b, @c, @d = line.split.map {|i| i.to_i }
    @cache = { 0 => 0 }
  end

  def pay(n = @n, lv = 0)
    return @cache[n] if @cache.has_key?(n)
    nn2, nm2 = n.divmod(2)
    p2 = @a + nm2 + nm2 * @d
    r2 = [pay(nn2, lv + 1) + p2, nn2, p2, "A + #{nm2}D"]
    nn3, nm3 = n.divmod(3)
    if nm3 > 3 - nm3
      nm3 = 3 - nm3
      nn3 += 1
    end
    p3 = @b + nm3 * @d
    r3 = [pay(nn3, lv + 1) + p3, nn3, p3, "B + #{nm3}D"]
    nn5, nm5 = n.divmod(5)
    if nm5 > 5 - nm5
      nm5 = 5 - nm5
      nn5 += 1
    end
    p5 = @c + nm5 * @d
    r5 = [pay(nn5, lv + 1) + p5, nn5, p5, "C + #{nm5}D"]
    p1 = n * @d
    r1 = [p1, n, p1, "#{n}D"]
    r = [r2, r3, r5, r1].min do |x, y|
      x[0] <=> y[0]
    end
    #$stderr.puts "#LV#{lv} #{n} -> #{r[1]} / pay #{r[3]} = #{r[2]}"
    return @cache[n] = r[0]
  end
end

T = $stdin.gets.to_i
0.upto(T - 1) do |i|
  e = Entry.new($stdin.gets)
  p = e.pay
  puts p
  #$stderr.puts "##cache = #{e.cache.length}"
end