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 Kotlin (1.3.71) (judge verdict: Wrong Answer)

fun main(args: Array<String>) {
  val n = readLine()!!.toInt()
  val a = Array(n){readLine()!!.split(' ').map(String::toDouble)}
  var num = 0.0
  var chk = 0.0
  var dif = 0.0
  var pay = 0.0
  var dct = 0
  var ect = 0
  var fct = 0
  for(i in 0..n-1){
    var d = a[i][0]
    var e = a[i][0]
    var f = a[i][0]
    while(d > 1.0){
      d = d / 2.0
      dct++
    }
    while(e > 1.0){
      e = e / 3.0
      ect++
    }
    while(f > 1.0){
      f = f / 5.0
      fct++
    }
    for(j in 0..dct){
      for(k in 0..ect){
        for(l in 0..fct){
          num = Math.pow(2.0 , j.toDouble()) * Math.pow(3.0 , k.toDouble()) * Math.pow(5.0 , l.toDouble())
          if(num >= a[i][0]){
            if(j != 0){
              dif = Math.abs(num/2.0 - a[i][0])
              if(a[i][1]*(j-1) + a[i][2]*k + a[i][3]*l + (dif+1.0)*a[i][4] < pay || pay.toInt() <= 0){
                pay = a[i][1]*(j-1) + a[i][2]*k + a[i][3]*l + (dif+1.0)*a[i][4]
              }else{
              }
            }else{
            }
            if(k != 0){
              dif = Math.abs(num / 3.0 - a[i][0])
              if(a[i][1]*j + a[i][2]*(k-1) + a[i][3]*l + (dif+1.0)*a[i][4] < pay || pay.toInt() <= 0){
                pay = a[i][1]*j + a[i][2]*(k-1) + a[i][3]*l + (dif+1.0)*a[i][4]
              }else{
              }
            }else{
            }
            if(l != 0){
              dif = Math.abs(num / 5.0 - a[i][0])
              if(a[i][1]*j + a[i][2]*k + a[i][3]*(l-1) + (dif+1.0)*a[i][4] < pay || pay.toInt() <= 0){
                pay = a[i][1]*j + a[i][2]*k + a[i][3]*(l-1) + (dif+1.0)*a[i][4]
              }else{
              }
            }else{
            }
            dif = Math.abs(num - a[i][0])
            if(a[i][1]*j + a[i][2]*k + a[i][3]*l + (dif+1.0)*a[i][4] < pay || pay.toInt() <= 0){
              pay = a[i][1]*j + a[i][2]*k + a[i][3]*l + (dif+1.0)*a[i][4]
            }else{
            }
            break
          }else{
          }
        }
      }
    }
    
    println(pay.toLong())
    pay = 0.0
    dct = 0
    ect = 0
    fct = 0
  }
}