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 <stdio.h>

int min(float a[]);

int main (void){
    int t;
    scanf("%d",&t);
    int n,a[4],ans=0;
    int b[3] = {2,3,5};
    float cost[4];
    int x=0;
    int minnum=0;

   // for(int i = 0;i <t;i++){
        scanf("%d %d %d %d %d",&n,&a[0],&a[1],&a[2],&a[3]);
        ans += a[3];//最初の一枚
        x ++;

        while(1){
            if(n ==x)break;
            if(n < x){
                ans += a[3]*(x-n);
                break;
            }
            cost[0] = (float)a[0]/(float)(b[0]*x-1);
            cost[1] = (float)a[1]/(float)(b[1]*x-1);
            cost[2] = (float)a[2]/(float)(b[2]*x-1);
            cost[3] = (float)a[3];
            minnum = min(cost);
            printf("%d\n",minnum);
            if(minnum==3){
                x ++;
                ans += a[3];
            }else{
                x *= b[minnum];
                ans += a[minnum];
            }
        }
        printf("%d\n",ans);
   // }

}

int min(float a[]){
    int min = 0;
    float m = a[0];
    printf("%f\n",a[0]);
    for(int i = 1;i<4;i++){
        printf("%f\n",a[i]);
        if(a[i] < m){
                 min = i;
                 m = a[i];
        }
    }
    return min;
}