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)

// Marcin Knapik
#include<bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")

using namespace std;

// #include <ext/pb_ds/assoc_container.hpp> // order_of_key
// #include <ext/pb_ds/tree_policy.hpp> // find by order
// using namespace __gnu_pbds; 
// #define ordered_set tree< int , null_type,less< int >, rb_tree_tag,tree_order_statistics_node_update>  

// const ll MAX_LL = 0xFFFFFFFFFFFFFFFF;
// mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define losuj(a, b) uniform_int_distribution<ll>(a, b)(rng)

typedef long double             ld;
typedef long long               ll;
typedef pair<int, int>          ii;
typedef vector<int>             vi;
typedef vector<ll>              vll;
typedef vector<vll>             vvll;
typedef vector<ii>              vii;
typedef vector<vi>              vvi;
typedef vector<vii>             vvii;
#define pll pair<long long, long long>

#define sz(a)                   (int)(a).size()
#define pb                      push_back
#define all(c)                  (c).begin(), (c).end()
#define rall(c)                 (c).rbegin(), (c).rend()
#define FOR(i, a)               for (int i = 0; i < (a); i++)
#define f                       first
#define s                       second

template <class T> inline bool setmin(T &a, T b){if (a > b)return a = b, 1;return 0;}
template <class T> inline bool setmax(T &a, T b){if (a < b)return a = b, 1;return 0;}

template<class T> inline T fast(T a,T b,T mod) {ll res = 1; while(b){if(b&1) res = (res*a)%mod;a = (a*a)%mod;b >>= 1;}return res;}
template<class T> istream &operator>>(istream &os, vector<T> &container){for (auto &u : container)os >> u;return os;}
template<class T> ostream &operator<<(ostream &os, const vector<T> &container){for (auto &u : container)os << u << " ";return os;}
template<class T> inline T gcd(T a, T b) { while (b)swap(a %= b, b); return a; }

const ll INF = 1e9 + 7;
const ll mod = 998244353;
const ll BIG_INF = 1e18 + 7;
const ll N = 1e6+7;
const ll T = 1<<20;
const ld inf = 1e18;

ll n, a, b, c, d;

ll dziel[] = {2, 3, 5};
ll koszt[3];
map<ll, ll> mapa;

ll req(ll x){
    if(x == 0)
        return 0;
    if(x == 1)
        return d;
    if(mapa.count(x))
        return mapa[x];
    ll ret = BIG_INF;
    if(ret < INF)
        ret = ret * d;

    FOR(i, 3){
        ll _ = ((x / dziel[i]) * dziel[i]);
        ret = min(ret, koszt[i] + req(_ / dziel[i]) + d * (x - _));
        _ += dziel[i];
        if(_ / dziel[i] != x)
            ret = min(ret, koszt[i] + req(_ / dziel[i]) + d * (_ - x));
    }
    return mapa[x] = ret;
}

signed main(){
    ios::sync_with_stdio(0);
    cin.tie(0);

    int q;
    cin >> q;
    while(q--){
        mapa.clear();
        cin >> n >> koszt[0] >> koszt[1] >> koszt[2] >> d;

        cout << req(n) << '\n';
    }
}