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>
using namespace std;
//using Graph = vector<vector<edge>>;
using dou =long double;
string yes="yes";
string Yes="Yes";
string YES="YES";
string no="no";
string No="No";
string NO="NO";

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
#define rep(i, n)         for(int i = 0; i < (int)(n); i++)
#define brep(n)           for(int bit=0;bit<(1<<n);bit++)
#define erep(i,container) for (auto i : container)
#define irep(i, n)        for(int i = n-1; i >= (int)0; i--)
#define rrep(i,m,n) for(ll i = m; i < (ll)(n); i++)
#define reprep(i,j,h,w) rep(i,h)rep(j,w)
#define all(x) (x).begin(),(x).end()
#define VEC(type,name,n) std::vector<type> name(n);rep(i,n)std::cin >> name[i];
#define pb push_back
#define pf push_front
#define query int qq;std::cin >> qq;rep(qqq,qq)
#define lb lower_bound
#define ub upper_bound
#define fi first
#define se second
#define itn int
#define mp make_pair
#define sum(a) accumulate(all(a),0ll)
#define keta fixed<<setprecision
#define vvector(name,typ,m,n,a)vector<vector<typ> > name(m,vector<typ> (n,a))
#define vvvector(name,t,l,m,n,a) vector<vector<vector<t> > > name(l, vector<vector<t> >(m, vector<t>(n,a)));
#define vvvvector(name,t,k,l,m,n,a) vector<vector<vector<vector<t> > > > name(k,vector<vector<vector<t> > >(l, vector<vector<t> >(m, vector<t>(n,a)) ));
#define case std::cout <<"Case #" <<qqq+1<<": "
#define res resize
#define as assign
#define ffor for(;;)
#define ppri(a,b) std::cout << a<<" "<<b << std::endl
#define pppri(a,b,c) std::cout << a<<" "<<b <<" "<< c<<std::endl
#define aall(x,n) (x).begin(),(x).begin()+(n)
#define ssum(a) accumulate(a,0ll) 

typedef long long ll;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;
//typedef long long T;
const int INF = 1000000000;
const ll INF64 = 9223372036854775807ll;
const ll MOD = 1000000007ll;
const dou pi=3.141592653589793;
//mapは、if関数などで覗いただけで構成されるので注意しよう！
ll n,a,b,c,d;
ll ans=INF64;
std::map<ll, ll> m;
int cou=0;
void dfs(ll num,ll val){
    if(m.find(num)!=m.end()){
        if(m[num]<=val)return;
        else chmin(m[num],val);
    }
    else{
        m[num]=val;
    }
    /*
    cou++;
    if(cou%20000==0){
        ppri(m.size(),ans);
    }
    */
    if(num<INF&&double(val)+double(num)*double(d)<double(INF64))chmin(ans,val+num*d);
    if(num==1)return;
    dfs(num/2,val+d*(num%2)+a);
    if(!num%2)dfs((num+1)/2,val+d*(2-(num%2))+a);
    //if(num<=2)return;
    dfs(num/3,val+d*(num%3)+b);
    if(!num%3)dfs((num+2)/3,val+d*(3-(num%3))+b);
    //if(num<=4)return;
    dfs(num/5,val+d*(num%5)+c);
    if(!num%5)dfs((num+4)/5,val+d*(5-(num%5))+c);
    return;
}

int main(){
    query{
        std::cin >> n>>a>>b>>c>>d;
        ans=INF64;
        m.clear();
        dfs(n,0ll);
        std::cout << ans << std::endl;
    }
}