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.h>
#include <algorithm>
#include <cstring>
#include <queue>
#include <assert.h>
#include <cmath>
#include <deque>
#include <unordered_map>
using namespace std;
using ll = long long;
using ld = long double;
const ll inf = 1000000000000000008;
ll N, A, B, C, D;
vector<ll> mul = {2, 3, 5};

unordered_map<ll, ll> Map;

ll calc(ll n) {
	if(n==0ll) return 0ll;
	if(n==1ll) return D;
	if(Map.count(n)) return Map[n];
	ll ret = (n<=1e9)? n*D:inf;
	for(auto x:mul) {
		ll ex;
		if(x==2) ex = A;
		else if(x==3) ex =B;
		else ex = C;
		if(n%x==0) ret = min(ret, calc(n/x)+ex);
		else {
			ll up = (n/x+1ll)*x;
			ll down = (n/x)*x;
			ret = min(ret, calc(down/x)+D*(n-down)+ex);
			ret = min(ret, calc(up/x)+D*(up-n)+ex);
		}
	}
	return Map[n] = ret;
}

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int tc; cin>>tc;
	while(tc--) {
		cin>>N>>A>>B>>C>>D;
		cout << calc(N) << '\n';
		Map.clear();
	}
	return 0;
}