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 <string.h>

int goal,x2,x3,x5,plus1;
int loop(int* db,unsigned long long int now);

int main(void){
    int db[200000]={0};
    int T;
    scanf("%d",&T);
    
    for(int i=0;i<T;i++){
        scanf("%d %d %d %d %d",&goal,&x2,&x3,&x5,&plus1);
        memset(db, 0, sizeof(db));
        db[1]=plus1;
        loop(db,1);
        printf("%d\n",db[goal]);
        for(int i=0;i<goal*2;i++){
            //printf("%d %llu\n",i,db[i]);
        }
    }
    
}

int loop(int* db,unsigned long long int now){
    if(now>goal*2){return 0;}
    //printf("%llu %llu\n",now,db[now]);
    if(now*2<goal*2){
    if(db[now*2]>db[now]+x2 || db[now*2]==0){
        db[now*2]=db[now]+x2;
        //printf("x2 %llu %llu %llu\n",now,now*2,db[now*2]);
        loop(db,now*2);
    }
    }
    
    if(now*3<goal*2){
    if(db[now*3]>db[now]+x3 || db[now*3]==0){
        db[now*3]=db[now]+x3;
        //printf("x3 %llu %llu %llu\n",now,now*3,db[now*3]);
        loop(db,now*3);
    }
    }
    
    if(now*5<goal*2){
    if(db[now*5]>db[now]+x5 || db[now*5]==0){
        db[now*5]=db[now]+x5;
        //printf("x5 %llu %llu %llu\n",now,now*5,db[now*5]);
        loop(db,now*5);
    }
    }
    
    if(now!=1){
    if(db[now-1]>db[now]+plus1|| db[now-1]==0){
    db[now-1]=db[now]+plus1;
    //printf("-1 %llu %llu %llu\n",now,now-1,db[now-1]);
    loop(db,now-1);
    }
    }
    
    if(db[now+1]>db[now]+plus1|| db[now+1]==0){
    db[now+1]=db[now]+plus1;
    //printf("+1 %llu %llu %llu\n",now,now+1,db[now+1]);
    loop(db,now+1);
    }
    
    return 0;
}