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>
#include <stdlib.h>
#include <string.h>
#define MOD 1000000007
#define INF 20000000000000
#define PI 3.14159265359
#define ull unsigned long long

int main(){
    ull t,n[11],a[11],b[11],c[11],d[11],i,num=0,coin=0,j,hyouka[3]={},tmp=0;
    scanf("%lld",&t);
    for (i=0; i<t; i++) scanf("%lld%lld%lld%lld%lld",&n[i],&a[i],&b[i],&c[i],&d[i]);
    for (j=0; j<t; j++) {
        num=0;coin=0;tmp=n[j];
        while (num!=n[j]) {
            if (num==0){
                num++;
                coin+=d[j];
            }
            else if(num==n[j]-1){
                num++;
                coin+=d[j];
            }
            else if(num==n[j]+1){
                num--;
                coin+=d[j];
            }else{
                hyouka[0]=(tmp/5)*c[j];
                hyouka[1]=(tmp/3)*b[j];
                hyouka[2]=(tmp/2)*a[j];
                if (hyouka[0]==0) hyouka[0]=INF;
                if (hyouka[1]==0) hyouka[1]=INF;
                if (hyouka[2]==0) hyouka[2]=INF;
                if (hyouka[0]<hyouka[1]&&hyouka[0]<hyouka[2]){num*=5;tmp/=5;coin+=c[j];}
                else if (hyouka[1]<hyouka[0]&&hyouka[1]<hyouka[2]){num*=3;tmp/=3;coin+=b[j];}
                else if (hyouka[2]<hyouka[1]&&hyouka[2]<hyouka[0]){num*=2;tmp/=2;coin+=a[j];}
                else {num++;coin++;}
            }
        }
        printf("%lld\n",coin);
    }
    
    return 0;
}