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;
typedef long long ll;
map<ll,ll>mp;
int a,b,c,d;
ll f(ll n)
{
    if(n==0)return 0;
    if(n==1)return d;
    if(mp.find(n)!=mp.end())return mp[n];
    ll res=2e18;
    if(!(n%2))res=f(n>>1)+a;
    else res=min(f(n>>1)+a+d,f((n+1)>>1)+a+d);
    int t=n%3;
    res=min(res,f(n/3)+b+1ll*d*t);
    if(t)res=min(res,f((n+2)/3)+b+1ll*d*(3-t));
    t=n%5;
    res=min(res,f(n/5)+c+1ll*d*t);
    if(t)res=min(res,f((n+4)/5)+c+1ll*d*(5-t));
    if(1.0*n*d<=1e18)res=min(res,n*d);
    mp[n]=res;
    return res;
}
int main()
{
    int _;
    scanf("%d",&_);
    while(_--)
    {
        ll n;
        scanf("%lld%d%d%d%d",&n,&a,&b,&c,&d);
        mp.clear();
        printf("%lld\n",f(n));
    }
    return 0;
}