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 long long ll;
ll read(){
	ll a=0,b=getchar(),c=1;
	while(!isdigit(b))c=b=='-'?-1:1,b=getchar();
	while(isdigit(b))a=a*10+b-'0',b=getchar();
	return a*c;
}
ll a,b,c,d,inf=9223372036854775807;
vector<ll> v;
map<ll,ll> m;
ll gao(ll n){
	if(n==0)return 0;
	if(n==1)return d;
	ll u=inf;
	if(n<inf/d)u=n*d;
	if(m.find(n)!=m.end())
		return m[n];
	ll x;
	for(ll i=0;i<3;i++){
		if(i==0)x=a;
		if(i==1)x=b;
		if(i==2)x=c;
		if(n%v[i]==0)
			u=min(u,x+gao(n/v[i]));
		else{
			ll y=n/v[i]*v[i];
			u=min(u,d*(n-y)+x+gao(n/v[i]));
			ll z=y+v[i];
			u=min(u,d*(z-n)+x+gao(n/v[i]+1));
		}
	}
	return m[n]=u;
}
int main(){
	v.push_back(2);
	v.push_back(3);
	v.push_back(5);
	ll t=read();
	while(t--){
		ll n=read();
		a=read(),b=read(),c=read(),d=read();
		m.clear();
		printf("%lld\n",gao(n));
	}
	return 0;
}