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>
#define LL long long
using namespace std;
const LL INF=0x7FFFFFFFFFFFFFFF;
map <LL,LL> M;
int G,A,B,C,D;
LL n;
LL calc(LL,int,int);
LL finds(LL num)
{
	if (M[num]) return M[num];
	LL res=INF;
	if (log10(num)+log10(D)<=18) res=num*D;
	if (num>=2) res=min(res,calc(num,2,A)); 
	if (num>=3) res=min(res,calc(num,3,B)); 
	if (num>=5) res=min(res,calc(num,5,C));
	return M[num]=res;
}
LL calc(LL num,int k,int cost)
{
	LL L=num/k*k,R=L+k,res=finds(L/k)+abs(num-L)*D+cost;
	if (L!=k) res=min(res,finds(R/k)+abs(num-R)*D+cost);
	return res;
}
int main()
{
	scanf("%d",&G);
	while (G--)
	{
		scanf("%lld%d%d%d%d",&n,&A,&B,&C,&D);
		M.clear();
		printf("%lld\n",finds(n));
	}
	return 0;
}