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;
using ll = long long;
using ull = unsigned long long;
#define REP(i,n) for (ll i = 0; i < (n); ++i)
#define RREP(i, n) for (ll i = (n) - 1; i >= 0; --i)
#define ALL(v) (v).begin(), (v).end()

ll a, b, c, d;
unordered_map<ll, ll> cache;
ll f(ll n){
    if(n == 0){ return 0; }
    if(n == 1){ return d; }
    if(cache.count(n)){ return cache[n]; }

    ll ret = numeric_limits<ll>::max();

    // div 2
    ret = min(
        ret,
        min(
            f((n + 1) / 2) + a + d * (n % 2),
            f(n / 2) + a + d * (n % 2)
        )
    );

    // div 3
    ret = min(
        ret,
        min(
            f((n + 2) / 3) + b + d * (3 - n % 3),
            f(n / 3) + b + d * (n % 3)
        )
    );

    // div 5
    ret = min(
        ret,
        min(
            f((n + 4) / 5) + c + d * (5 - n % 5),
            f(n / 5) + c + d * (n % 5)
        )
    );

    if(ret / d > n){
        ret = min(ret, n * d);
    }
    cache[n] = ret;
    return ret;
}

int main(){
    ll t;
    cin >> t;
    REP(i, t){
        ll n;
        cin >> n >> a >> b >> c >> d;
        cache.clear();
        cout << f(n) << endl;
    }
    return 0;
}