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: Runtime Error)

#include<stdio.h>
#include<stdlib.h>

long long int min(long long int,long long int y,long long int z,long long int w);

int main(void){
	long long int t,n,a,b,c,d;
	scanf("%lld",&t);
	long long int *ans;
	ans=(long long int*)malloc(10*sizeof(long long int));
	
	for(int j=0;j<t;j++){
		long long int i,m;
		scanf("%lld %lld %lld %lld %lld",&n,&a,&b,&c,&d);
		m=(2*a)/(3*d)+(4*n)/3 > (3*b)/(4*d)+(3*n)/2 ? (2*a)/(3*d)+(4*n)/3 : (3*b)/(4*d)+(3*n)/2;
		m= m > (5*c)/(6*d)+(5*n)/3 ? m : (5*c)/(6*d)+(5*n)/3;
		m+=3;
		long long int *cost; /*最小値をdpで求めたい*/
		cost=(long long int*)malloc(m*sizeof(long long int));
		cost[n]=0;
		
		for(i=n+1;i<m;i++){
			cost[i]=cost[i-1]+d;
			//printf("%lld %lld\n",i,cost[i]);
		}
		
		for(i=n-1;i>0;i--){
			if(5*i>=m&&3*i>=m&&2*i>=m){
				cost[i]=cost[i+1]+d;
			}else if(5*i>=m&&3*i>=m){
				cost[i]=(cost[2*i]+a)<(cost[i+1]+d) ? cost[2*i]+a : cost[i+1]+d;
				//printf("%lld %lld",cost[i+1],(cost[i+1]+c));
			}else if(5*i>=m){
				cost[i]=(cost[2*i]+a)<(cost[i+1]+d) ? cost[2*i]+a : cost[i+1]+d;
				if(cost[3*i]+b<cost[i]){
					cost[i]=cost[3*i]+b;
				}
			}else{
				cost[i]=min(cost[2*i]+a,cost[3*i]+b,cost[5*i]+c,cost[i+1]+d);
			}
			
			//printf("%lld %lld\n",i,cost[i]);
		
		}
		ans[j]=cost[1]+d;
		free(cost);
	}
	for(int j=0;j<t;j++){
		printf("%lld\n",ans[j]);
	}
}

long long int min(long long int x,long long int y,long long int z,long long int w){
	long long int temp1,temp2,ret;
	temp1=x<y?x:y;
	temp2=z<w?z:w;
	ret=temp1<temp2?temp1:temp2;
	return ret;
}