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 C (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>

#define HASH 100003

typedef struct List {
	struct List* next;
	unsigned long long v;
} list;

typedef struct {
	unsigned long long key, id;
} data;

typedef struct {
	data obj[1000000];
	int size;
} min_heap;

void push(data x, min_heap* h)
{
	int i = h->size;
	data tmp;
	h->obj[(h->size)++] = x;
	while (1) {
		if (i == 0) break;
		else if (h->obj[i].key < h->obj[(i-1)/2].key) {
			tmp = h->obj[(i-1)/2];
			h->obj[(i-1)/2] = h->obj[i];
			h->obj[i] = tmp;
			i = (i - 1) / 2;
		} else break;
	}
}

data pop(min_heap* h)
{
	int i = 0, j;
	data output = h->obj[0], tmp;
	h->obj[0] = h->obj[--(h->size)];
	while (1) {
		if (i * 2 + 1 >= h->size) break;
		else if (i * 2 + 2 >= h->size) j = i * 2 + 1;
		else if (h->obj[i*2+1].key < h->obj[i*2+2].key) j = i * 2 + 1;
		else j = i * 2 + 2;
		if (h->obj[j].key < h->obj[i].key) {
			tmp = h->obj[j];
			h->obj[j] = h->obj[i];
			h->obj[i] = tmp;
			i = j;
		} else break;
	}
	return output;
}

int main()
{
	int i, j, k, T;
	unsigned long long N, A, B, C, D, tmp, cost;
	const unsigned long long sup = 0xffffffffffffffff;
	list *p, *hash[HASH], e[1000000];
	data d;
	min_heap h;
	scanf("%d", &T);
	for (i = 1; i <= T; i++) {
		scanf("%llu %llu %llu %llu %llu", &N, &A, &B, &C, &D);
		
		for (j = 0, k = 0; j < HASH; j++) hash[j] = NULL;
		h.size = 0;
		d.key = 0;
		d.id = N;
		push(d, &h);
		
		while (h.obj[0].id > 0) {
			d = pop(&h);
			j = d.id % HASH;
			for (p = hash[j]; p != NULL; p = p->next) if (p->v == d.id) break;
			if (p != NULL) continue;
			
			e[k].v = d.id;
			e[k].next = hash[j];
			hash[j] = &(e[k++]);
			tmp = d.id;
			cost = d.key;
			
			if (tmp <= (sup - cost) / D) {
				d.id = 0;
				d.key = cost + D * tmp;
				push(d, &h);
			}
			
			for (j = 0; j <= 1; j++) {
				if ((tmp + j) % 2 == 0) {
					d.id = (tmp + j) / 2;
					d.key = cost + A + D * j;
					push(d, &h);
					if (tmp > j && j != 0) {
						d.id = (tmp - j) / 2;
						push(d, &h);
					}
				}
			}
			
			for (j = 0; j <= 2; j++) {
				if ((tmp + j) % 3 == 0) {
					d.id = (tmp + j) / 3;
					d.key = cost + B + D * j;
					push(d, &h);
				}
				if (tmp > j && j != 0 && (tmp - j) % 3 == 0) {
					d.id = (tmp - j) / 3;
					d.key = cost + B + D * j;
					push(d, &h);
				}
			}
			
			for (j = 0; j <= 4; j++) {
				if ((tmp + j) % 5 == 0) {
					d.id = (tmp + j) / 5;
					d.key = cost + C + D * j;
					push(d, &h);
				}
				if (tmp > j && j != 0 && (tmp - j) % 5 == 0) {
					d.id = (tmp - j) / 5;
					d.key = cost + C + D * j;
					push(d, &h);
				}
			}
		}
		printf("%llu\n", h.obj[0].key);
	}
}