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>

#define NUM 100000000000

int main(void){

        int a[NUM],b[NUM],c[NUM],d[NUM],t;
        int n[NUM];

        scanf("%d",&t);

        int coin[10000] = {0};

for(int i=0;i<t;i++){
        scanf("%d%d%d%d%d",&n[i],&a[i],&b[i],&c[i],&d[i]);

while(n[i]>0){

if(n[i]%5 == 0){ 
        n[i] = n[i]/5;
        coin[i] = coin[i] + a[i];
}else if(n[i]%3 == 0){ 
        n[i] = n[i]/3;
        coin[i] = coin[i] + b[i];
}else if(n[i] % 2 == 0){ 
        n[i] = n[i]/2;
        coin[i] = coin[i] + c[i];
}else{
        n[i] = n[i] - 1;
        coin[i] = coin[i] + d[i];
}
}
printf("%d\n",coin[i]);
}
return 0;
}