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: Accepted)

#include <bits/stdc++.h>
using namespace std;
  
#define int long long 
#define pb push_back
#define S second
#define F first
#define f(i,n) for(int i=0;i<n;i++)
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define vi vector<int>
#define pii pair<int,int>

const int inf = 1e15;

int a,b,c,d,n;

map<int,int> mp;

int solve(int n)
{
    if(n == 0) return inf;
    if(n == 1) return 0;
    
    if(mp.find(n) != mp.end()) return mp[n];
    
    int res = inf;
    
    if(d < 10000 && n < 1e12) res = min(res,(n-1)*d);
    
    if(n%2 == 0) res = min(res,solve(n/2) + a);
    else res = min(res,  min( solve((n-1)/2) + a + d, solve((n+1)/2)+ a + d ));
    
    if(n%3 == 0) res = min(res,solve(n/3) + b);
    else if(n%3 == 1) res = min(res,solve((n-1)/3) + d + b);
    else res = min(res,solve((n+1)/3) + d + b);
        
    if(n%5 == 0) res = min(res,solve(n/5) + c);
    else if(n%5 == 1) res = min(res,solve(n/5) + c + d);
    else if(n%5 == 4) res = min(res,solve((n+1)/5) + c + d);
    else if(n%5 == 3) res = min(res,solve((n+2)/5) + c + d + d);
    else res = min(res, solve(n/5) + c + d + d);
    
    mp[n] = res;
        
    return res;
}

signed main()
{
    fast;
    
    int t = 1;
    
    cin >> t;
    
    while(t--)
    {
        mp.clear();
        
        cin >> n >> a >> b >> c >> d;
        cout << solve(n) + d<< '\n';
    }
}