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: Time Limit Exceeded)

#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;

int factorization(int64_t n, int a){
    int m=0;
    while(n%a==0){
        n /= a;
        m++;
    }
    return m;
}

int main(){
    int T;
    cin >> T;
    int64_t n, a, b, c, d, minimum;

    for(int t=0;t<T;t++){
        cin >> n >> a >> b >> c >> d;
        if(n==1){
            cout << d << endl;
            continue;
        }
        int64_t nn, aa, bb, cc, count, N;
        count = 0;
        for(int64_t i=-pow(10,5);i<pow(10,5);i++){
            nn = n + i;

            if(nn<0 || (n%2!=0 && n%3!=0 && n%5!=0)){
                continue;
            }

            aa = factorization(nn,a);
            bb = factorization(nn,b);
            cc = factorization(nn,c);

            N = aa*a + bb*b + cc*c + (abs(i)+1)*d;

            if(count == 0){
                minimum = N;
            }else{
                minimum = min(N,minimum);
            }
        }
        cout << minimum << endl;
    }

    return 0;
}