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)



import java.util.*
import kotlin.math.min

fun main() {

    val coins = arrayOf(2L,3L,5L) //61, 39, 27

    val sc = Scanner(System.`in`)
    val t = sc.nextInt()

    repeat(t) {
        val n = sc.nextLong()
        val costs = LongArray(3)
        costs[0] = sc.nextLong()
        costs[1] = sc.nextLong()
        costs[2] = sc.nextLong()

        val d = sc.nextLong()
        val cache = HashMap<Long,Long>()

        fun bestCost(v:Long, c:Int, p : Long, r:Long): Long {
            if(p == 0L){
                return Long.MAX_VALUE
            }
            val remainder = r * d
            val cost = costs[c] + remainder
            val diff = v - p
            val costD = diff * d
            if(costD < 0)
                return cost
            return min(costD, cost)
        }

        fun solve(remain:Long): Long {
            if(remain == 0L)
                return 0L
            if(remain == 1L)
                 return d
            if(cache.contains(remain)) {
                return cache[remain]!!
            }
            var best = Long.MAX_VALUE
            for(i in coins.indices) {
                val c = coins[i]
                val p = remain / c
                val r = remain % c
                val up = p + 1L
                val ur = c - r
                best = min(best, bestCost(remain, i, p, r) + solve(p))
                if(ur == c) {
                    continue
                }
                best = min(best, bestCost(remain, i, up, ur) + solve(up))
            }
            cache[remain] = best
            return best
        }
        println(solve(n))


    }


}