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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,ini,n) for(int i=ini;i<n;i++)
#define _rep(i,ini,n) for(int i=ini;i>=n;i--)
#define ToEnd(a) a.begin(),a.end()
uint64_t MOD=1000000007;

int64_t N,A,B,C,D;
map<int64_t,int64_t> mp;
int64_t ans=INT64_MAX;

void wfs(queue<pair<int64_t,int64_t>> &q,map<int64_t,int64_t> &m,int64_t n,int64_t c){
    if(m.count(n)) m[n]=min(m[n],c);
    else  m[n]=c;
    //cout<<n<<" "<<c<<endl;

    if(n==1){
        ans=min(ans, c+D);
        return;
    }

    if(n%2==0 && (!m.count(n/2) || c+A<m[n/2])) q.push(make_pair(n/2,c+A));
    if(n%3==0 && (!m.count(n/3) || c+B<m[n/3])) q.push(make_pair(n/3,c+B));
    if(n%5==0 && (!m.count(n/5) || c+C<m[n/5])) q.push(make_pair(n/5,c+C));

    if(!m.count(n-1) || c+D<m[n-1]) q.push(make_pair(n-1,c+D));
    if((n+1)<=N+30-N%30 && (!m.count(n+1) || c+D<m[n+1])) q.push(make_pair(n+1,c+D));
}

int main(){
    int T; cin>>T;
    rep(t,0,T){
        cin>>N>>A>>B>>C>>D;

        mp.clear();
        ans=INT64_MAX;
        queue<pair<int64_t,int64_t>> q;
        q.push(make_pair(N,0));

        while(!q.empty()){
            int64_t n,c;
            tie(n,c)=q.front();
            q.pop();

            wfs(q,mp,n,c);
        }
        cout<<ans<<endl;
    }
}