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)

t = gets.to_i

def rec(n, memo, a, b, c, d)
  return 0 if n==0
  return d if n==1
  return memo[n] if memo[n]
  
  min = n*d
  if n%2==0
    tmp = rec(n/2, memo, a, b, c, d)+a
    min = tmp if min > tmp
  else
    m = n%2
    l = 2-m
    tmp1 = rec((n-m)/2, memo, a, b, c, d)+a+(d*m)
    tmp2 = rec((n+l)/2, memo, a, b, c, d)+a+(d*l)
    min = tmp1 if min > tmp1
    min = tmp2 if min > tmp2
  end
  
  if n%3==0
    tmp = rec(n/3, memo, a, b, c, d)+b
    min = tmp if min > tmp
  else
    m = n%3
    l = 3-m
    tmp1 = rec((n-m)/3, memo, a, b, c, d)+b+(d*m)
    tmp2 = rec((n+l)/3, memo, a, b, c, d)+b+(d*l)
    min = tmp1 if min > tmp1
    min = tmp2 if min > tmp2
  end
  
  if n%5==0
    tmp = rec(n/5, memo, a, b, c, d)+c
    min = tmp if min > tmp
  else
    m = n%5
    l = 5-m
    tmp1 = rec((n-m)/5, memo, a, b, c, d)+c+(d*m)
    tmp2 = rec((n+l)/5, memo, a, b, c, d)+c+(d*l)
    min = tmp1 if min > tmp1
    min = tmp2 if min > tmp2
  end
  
  #p [n,min]
  memo[n] = min if !memo[n]
  return min
end

t.times{
  n,a,b,c,d = gets.split.map(&:to_i)
  memo = {}
  p rec(n, memo, a, b, c, d)
}