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>

#define debug(var)  do{std::cerr << #var << " : ";view(var);}while(0)
template<typename T> void view(T e){std::cerr << e << std::endl;}
template<typename T> void view(const std::vector<T>& v){for(const auto& e : v){ std::cerr << e << " "; } std::cerr << std::endl;}
template<typename T> void view(const std::vector<std::vector<T> >& vv){ for(const auto& v : vv){ view(v); } }
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
using namespace std;

typedef long long ll;
typedef long double ld;
typedef vector<ll> v1;
typedef vector<v1> v2;
typedef vector<v2> v3;

typedef unordered_map<ll, unordered_map<ll, ll>> graph;

const ll INF = 1ll << 50;
const ll mod = 1000000007;

ll t;
ll n;
ll a,b,c,d;

void add(ll y, ll cost, unordered_map<ll,ll>& m, priority_queue<ll>& pq){

    if(m.find(y) == m.end()){
        m[y] = cost;
        pq.push(y);
    }else{
        chmin(m[y], cost);
    }
}

void move(ll x, ll cost, ll k, ll a, ll d, unordered_map<ll,ll>& m, priority_queue<ll>& pq){
    if(x%k == 0){
        ll y = x/k;
        add(y, cost+a, m, pq);
    }else{
        ll y = x/k;
        ll diff = abs(x-y*k);
        add(y, cost+a+d*diff, m, pq);
        y = x/k+1;
        diff = abs(x-y*k);
        add(y, cost+a+d*diff, m, pq);
    }
}

void solve(ll n, ll a, ll b, ll c, ll d){
    unordered_map<ll,ll> m;
    priority_queue<ll> pq;
    pq.push(n);
    m[n] = 0;
    ll ans = INF;
    while(!pq.empty()){
        ll x = pq.top();
        ll cost = m[x];
        pq.pop();
        // debug(x);
        // debug(cost);
        if(x < 2000000000) chmin(ans, cost+x*d);
        move(x, cost, 2, a, d, m, pq);
        move(x, cost, 3, b, d, m, pq);
        move(x, cost, 5, c, d, m, pq);

    }
    cout << ans << endl;


}

int main(){
    cin >> t;
    for(ll i = 0;i < t;i++){
        cin >> n >> a >> b >> c >> d;
        solve(n, a, b, c, d);
    }
}