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)

# 参考: https://atcoder.jp/contests/agc044/submissions/13654472

class TestCase
  def initialize(a, b, c, d)
    @a, @b, @c, @d = a, b, c, d
    @memo = {0 => 0, 1 => d}              # 初期値
  end
  
  def go(n)
    return @memo[n] if @memo.key? n
    ad, bd, cd = n/2, n/3, n/5
    ai, bi, ci = (n+1)/2, (n+2)/3, (n+4)/5
    # `n-bd*3` は `n%3` と同じ意味
    # `bi*3-n` は `(3-n%3)%3` と同じ意味
    @memo[n] ||= [
      @d * n,                             # ずっとD
      @a + @d * (n&1) + go(ad),           # Dで2の倍数に減らしてA
      @b + @d * (n-bd*3) + go(bd),        # Dで3の倍数に減らしてB
      @c + @d * (n-cd*5) + go(cd),        # Dで5の倍数に減らしてC
      @a + @d * (n&1) + go(ai),           # Dで2の倍数に増やしてA
      @b + @d * (bi*3-n) + go(bi),        # Dで2の倍数に増やしてB
      @c + @d * (ci*5-n) + go(ci)         # Dで2の倍数に増やしてC
      ].min
  end
end

gets.to_i.times do
  n, a, b, c, d = gets.chomp.split(" ").map(&:to_i)
  t = TestCase.new(a, b, c, d)
  puts t.go(n)
end

=begin
input:
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

output:
20
19
26
3821859835
23441258666
=end