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 Go (1.14.1) (judge verdict: Accepted)

package main

import (
	"bufio"
	"fmt"
	"log"
	"math"
	"os"
	"strconv"
)

var sc = bufio.NewScanner(os.Stdin)
var buff []byte

func init() {
	sc.Split(bufio.ScanWords)
	sc.Buffer(buff, bufio.MaxScanTokenSize*1024)
	log.SetFlags(log.Lshortfile)
}

var memo map[int]int

var A, B, C, D int

func solve(N int) int {
	if N == 0 {
		return 0
	}
	if N == 1 {
		return D
	}
	if v, ok := memo[N]; ok {
		// log.Println(N, v, ok)
		return v
	}
	l2 := (N / 2) * 2
	r2 := ((N + 1) / 2) * 2
	l3 := (N / 3) * 3
	r3 := ((N + 2) / 3) * 3
	l5 := (N / 5) * 5
	r5 := ((N + 4) / 5) * 5
	res := int(1e18)
	if N < res/D {
		res = N * D
	}
	res = min(res, abs(l2-N)*D+A+solve(l2/2))
	res = min(res, abs(r2-N)*D+A+solve(r2/2))
	res = min(res, abs(l3-N)*D+B+solve(l3/3))
	res = min(res, abs(r3-N)*D+B+solve(r3/3))
	res = min(res, abs(l5-N)*D+C+solve(l5/5))
	res = min(res, abs(r5-N)*D+C+solve(r5/5))
	memo[N] = res
	return res
}

func main() {
	T := nextInt()
	for i := 0; i < T; i++ {
		memo = map[int]int{}
		N := nextInt()
		A = nextInt()
		B = nextInt()
		C = nextInt()
		D = nextInt()
		fmt.Println(solve(N))
	}
}

func nextString() string {
	sc.Scan()
	return sc.Text()
}

func nextFloat64() float64 {
	sc.Scan()
	f, err := strconv.ParseFloat(sc.Text(), 64)
	if err != nil {
		panic(err)
	}
	return f
}

func nextInt() int {
	sc.Scan()
	i, err := strconv.Atoi(sc.Text())
	if err != nil {
		panic(err)
	}
	return i
}

func nextInts(n int) (r []int) {
	r = make([]int, n)
	for i := 0; i < n; i++ {
		r[i] = nextInt()
	}
	return r
}

var dy = []int{0, 1, 0, -1}
var dx = []int{1, 0, -1, 0}
var MAX = math.MaxInt32

func pow(a, b int) int {
	return int(math.Pow(float64(a), float64(b)))
}

func lcm(a, b int) int {
	return a * b / gcd(a, b)
}
func gcd(a, b int) int {
	if a < b {
		a, b = b, a
	}
	for b != 0 {
		a, b = b, a%b
	}
	return a
}
func max(a ...int) int {
	r := a[0]
	for i := 0; i < len(a); i++ {
		if r < a[i] {
			r = a[i]
		}
	}
	return r
}
func min(a ...int) int {
	r := a[0]
	for i := 0; i < len(a); i++ {
		if r > a[i] {
			r = a[i]
		}
	}
	return r
}
func sum(a []int) (r int) {
	for i := range a {
		r += a[i]
	}
	return r
}
func abs(a int) int {
	if a < 0 {
		return -a
	}
	return a
}

type Pair struct {
	a, b int
}

type Pairs []Pair

func (p Pairs) Len() int {
	return len(p)
}
func (p Pairs) Swap(i, j int) {
	p[i], p[j] = p[j], p[i]
}

func (p Pairs) Less(i, j int) bool {
	if p[i].a == p[j].a {
		return p[i].b < p[j].b
	}
	return p[i].a < p[j].a
}