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;


#define li          long long int
#define rep(i,to)   for(li i=0;i<((li)(to));i++)
#define repp(i,start,to)    for(li i=(li)(start);i<((li)(to));i++)
#define pb          push_back
#define sz(v)       ((li)(v).size())
#define bgn(v)      ((v).begin())
#define eend(v)     ((v).end())
#define allof(v)    (v).begin(), (v).end()
#define dodp(v,n)       memset(v,(li)n,sizeof(v))
#define bit(n)      (1ll<<(li)(n))
#define mp(a,b)     make_pair(a,b)
#define rin rep(i,n)
#define EPS 1e-12
#define ETOL 1e-8
#define MOD 1000000007
typedef pair<li, li> PI;

#define INF bit(60)

#define DBGP 1


#define idp if(DBGP)
#define F first
#define S second
#define p2(a,b)     idp cout<<a<<"\t"<<b<<endl
#define p3(a,b,c)       idp cout<<a<<"\t"<<b<<"\t"<<c<<endl
#define p4(a,b,c,d)     idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<endl
#define p5(a,b,c,d,e)       idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<endl
#define p6(a,b,c,d,e,f)     idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<endl
#define p7(a,b,c,d,e,f,g)       idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<endl
#define p8(a,b,c,d,e,f,g,h)     idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<"\t"<<h<<endl
#define p9(a,b,c,d,e,f,g,h,i)       idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<"\t"<<h<<"\t"<<i<<endl
#define p10(a,b,c,d,e,f,g,h,i,j)        idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<"\t"<<h<<"\t"<<i<<"\t"<<j<<endl
#define foreach(it,v)   for(__typeof((v).begin()) it=(v).begin(); it!=(v).end(); ++it)
#define p2p(x)      idp p2((x).F, (x).S)
#define dump(x,n)   idp{rep(i,n){cout<<x[i]<<" ";}puts("");}
#define dump2(x,n)  idp{rep(i,n){cout<<"["<<x[i].F<<" , "<<x[i].S<<"] ";}puts("");}
#define dumpi(x)    idp{foreach(it, x){cout<<(*it)<<" ";}puts("");}
#define dumpi2(x)   idp{foreach(it, x){cout<<"["<<(it)->F<<" , "<<(it)->S<<"] ";}puts("");}

#define read2d(a,w,h)   rep(i,h)rep(j,w)cin>>a[i][j]
#define dump2d(a,w,h)   rep(i,h){rep(j,w)cout<<a[i][j]<<" ";puts("");}

typedef pair<li, li> PI;

inline void solve() {
    li n, a, b, c, d;
    cin >> n >> a >> b >> c >> d;

    map<li, li> memo;

    // cost, pos
    priority_queue<PI> pq;

    auto try_push = [&](li cost, li prev, li prev_cost, li num) {
        if ((abs(prev - num)*d) / d == abs(prev - num) && prev_cost + abs(prev - num)*d >= 0) {
            cost = min(cost, prev_cost + abs(prev - num) * d);
        }
        if (memo.count(num) == 0 || memo[num] > cost) {
            pq.push({ -cost, num});
        }
    };
    pq.push({0, n});
    while (!pq.empty()) {
        li now = pq.top().S;
        li now_cost = -pq.top().F;
        pq.pop();
        if (memo.count(now) > 0 && memo[now] <= now_cost) {
            continue;
        }
        memo[now] = now_cost;
        //p2(now, now_cost);
        if (now == 0) {
            cout << now_cost << endl;
            return;
        }
        if (now % 2 == 0) {
            try_push(now_cost + a, now, now_cost, now / 2);
        } else {
            try_push(now_cost + a + d, now, now_cost,  now / 2);
            try_push(now_cost + a + d, now, now_cost,  now / 2 + 1);
        }
        if (now % 3 == 0) {
            try_push(now_cost + b, now, now_cost,  now / 3);
        } else {
            li amari = now % 3;
            try_push(now_cost + b + d * amari, now, now_cost,  now / 3);
            try_push(now_cost + b + d * (3 - amari), now, now_cost,  now / 3 + 1);
        }
        if (now % 5 == 0) {
            try_push(now_cost + c, now, now_cost,  now / 5);
        } else {
            li amari = now % 5;
            try_push(now_cost + c + d * amari, now, now_cost,  now / 5);
            try_push(now_cost + c + d * (5 - amari), now, now_cost,  now / 5 + 1);
        }
    }
}

int main() {
    li t;
    cin >> t;
    rep(i, t) {
        solve();
    }

    return 0;
}