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 <bits/stdc++.h>
using namespace std;
typedef vector<int> VI;
typedef pair <int,int> ii;
typedef long long LL;
#define pb push_back
const int INF = 2147483647;
const LL INFF = 1000000000000000000LL;
const int N = 105;

int val[3] = {2, 3, 5};
int cost[3];
map<LL, LL> mapa;
LL n;
int d, z;

LL go(LL n) {
	if (n == 1) return d;
	if (n == 0) return 0;
	if (mapa.find(n) != mapa.end()) return mapa[n];
	LL minn = INFF;
	if ((n * d) / d == n && n * d > 0) minn = min(minn, n * d);
	for (int i=0;i<3;i++) {
		int k = n % val[i];
		minn = min(minn, go(n / val[i]) + cost[i] + d * 1LL * k);
		if (k != 0) minn = min(minn, go(n / val[i] + 1) + cost[i] + d * 1LL * (val[i] - k));
	}
	mapa[n] = minn;
	return minn;
}

int main() {
scanf("%d", &z);
while (z--) {
	scanf("%lld %d %d %d %d", &n, &cost[0], &cost[1], &cost[2], &d);
	mapa.clear();
	printf("%lld\n", go(n));
}	
return 0;
}