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: Runtime Error)

import kotlin.math.*

fun main(){
    val t = readLine()!!.toInt()
    repeat(t){
        val (n, a, b, c, d) = readLine()!!.split(" ").map(String::toLong)
        val memo = mutableMapOf<Long, Long>()
        val array = arrayOf(Pair(2L, a), Pair(3L, b), Pair(5L, c))

        fun search(now: Long): Long{
            if(now == 0L){
                return 0
            }else if(now == 1L){
                return d
            }else if(memo.containsKey(now)){
                return memo[now]!!
            }else{
                var res = Long.MAX_VALUE
                if(now < res / d) res = now * d
                array.forEach {
                    val l = (now / it.first) * it.first
                    val r = ((now + it.first - 1L) / it.first) * it.first
                    res = min(res, d * abs(l - now) + it.second + search(l/it.first))
                    res = min(res, d * abs(r - now) + it.second + search(r/it.first))
                }
                memo[now] = res
                return res
            }
        }
        search(n)
        println(memo[n]!!)
    }
}