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: Wrong Answer)

#include <iostream>
#include <vector>
#include <string>
using namespace std;

using cat = long long;

cat A, B, C, D, ans;
constexpr int MX = 5000000;
constexpr cat rep = 1000000;
int cur_rep = 0;
vector<cat> ans_sm(MX+1);

void prec(cat N) {
	ans_sm[N] = D*N;
	if(N <= 1) return;
	cat d3 = N%3, d5 = N%5, n3 = N/3, n5 = N/5;
	if(d3 == 2) d3 = 1, n3++;
	if(d5 == 3) d5 = 2, n5++;
	if(d5 == 4) d5 = 1, n5++;
	ans_sm[N] = min(ans_sm[N], d5*D+C+ans_sm[n5]);
	ans_sm[N] = min(ans_sm[N], d3*D+B+ans_sm[n3]);
	cat d2 = N%2, n2 = N/2;
	ans_sm[N] = min(ans_sm[N], d2*D+A+ans_sm[n2]);
	if(d2) ans_sm[N] = min(ans_sm[N], d2*D+A+ans_sm[n2+1]);
}

void bt(cat N, cat cur) {
	if(N <= MX) {
		ans = min(ans, cur + ans_sm[N]);
		return;
	}
	if(N <= ans/D) ans = min(ans, cur+D*N);
	if(cur >= ans) return;
	if(cur_rep == rep) return;
	cur_rep++;
	cat d3 = N%3, d5 = N%5, n3 = N/3, n5 = N/5;
	if(d3 == 2) d3 = 1, n3++;
	if(d5 == 3) d5 = 2, n5++;
	if(d5 == 4) d5 = 1, n5++;
	bt(n5, cur+d5*D+C);
	bt(n3, cur+d3*D+B);
	cat d2 = N%2, n2 = N/2;
	bt(n2, cur+d2*D+A);
	if(d2) bt(n2+1, cur+d2*D+A);
}

int main() {
	cin.sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int T;
	cin >> T;
	while(T--) {
		cat N;
		cin >> N >> A >> B >> C >> D;
		for(int i = 0; i <= MX; i++) prec(i);
		cur_rep = 0;
		ans = 1000*(A+D);
		bt(N, 0);
		cout << ans << endl;
	}
}