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<bits/stdc++.h>
#include<unordered_set>
#include<unordered_map>
using namespace std;
#define int long long
#define REP(i,m,n) for(int i=(m);i<(n);i++)
#define rep(i,n) REP(i,0,n)
#define pb push_back
#define all(a) a.begin(),a.end()
#define rall(c) (c).rbegin(),(c).rend()
#define mp make_pair
#define endl '\n'
typedef long long ll;
typedef pair<ll,ll> pll;
typedef long double ld;
const ll inf=1e9+7;
const ll mod=1e9+7;
ll a,b,c,d;
map<ll,ll>ma;
ll solve(ll now){
    if(ma.find(now)!=ma.end()){
        return ma[now];
    }
    ll res;
    ll k;
    if(now>=inf){
        k=2;
        if(now%k==0){
            res=solve(now/k)+a;
        }else{
            res=solve(now/k)+a+now%k*d;
            res=min(solve(now/k+1)+a+(k-now%k)*d,res);
        }
    }else{
        res=now*d;
    }
    if(now/2>0){
        k=2;
        if(now%k==0){
            res=min(solve(now/k)+a,res);
        }else{
            res=min(solve(now/k)+a+now%k*d,res);
            res=min(solve(now/k+1)+a+(k-now%k)*d,res);
        }
    }
    if(now/3>0){
        k=3;
        if(now%k==0){
            res=min(solve(now/k)+b,res);
        }else{
            res=min(solve(now/k)+b+now%k*d,res);
            res=min(solve(now/k+1)+b+(k-now%k)*d,res);
        }
    }
    if(now/5>0){
        k=5;
        if(now%k==0){
            res=min(solve(now/k)+c,res);
        }else{
            res=min(solve(now/k)+c+now%k*d,res);
            res=min(solve(now/k+1)+c+(k-now%k)*d,res);
        }
    }
    ma[now]=res;
    return res;
}
signed main(){
    ll t;cin>>t;
    while(t){
        t--;
        ll n;cin>>n>>a>>b>>c>>d;
        ma.clear();
        ma[0]=0;
        ll ans=solve(n);
        cout<<ans<<endl;
    }
}