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;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
const ll INF = 9223372036854775807; //MAX 9223372036854775807

int main()
{
ll t ;
cin >> t ;
rep(i,t){
  ll n ;
  cin >> n ;
  ll a , b , c , d ;
  cin >> a >> b >> c >> d ;
  map<ll,ll> m;
  m[0]=INF;
  queue<ll> q;
  q.push(n);
  m[n]=1;
  while(!q.empty()){
    ll p=q.front();
    q.pop();
    if(m[p/2]==0||m[p/2]>m[p]+a+(p%2)*d ){
      m[p/2]=m[p]+a+(p%2)*d;
      q.push(p/2);
    }
    if(m[p/3]==0||m[p/3]>m[p]+b+(p%3)*d ){
      m[p/3]=m[p]+b+(p%3)*d;
      q.push(p/3);
    }
    if(m[p/5]==0||m[p/5]>m[p]+c+(p%5)*d ){
      m[p/5]=m[p]+c+(p%5)*d;
      q.push(p/5);
    }
    if(m[(p+1)/2]==0||m[(p+1)/2]>m[p]+a+(2-p%2)*d ){
      m[(p+1)/2]=m[p]+a+(2-p%2)*d;
      q.push((p+1)/2);
    }
    if(m[(p+2)/3]==0||m[(p+2)/3]>m[p]+b+(3-p%3)*d ){
      m[(p+2)/3]=m[p]+b+(3-p%3)*d;
      q.push((p+2)/3);
    }
    if(m[(p+4)/5]==0||m[(p+4)/5]>m[p]+c+(5-p%5)*d ){
      m[(p+4)/5]=m[p]+c+(5-p%5)*d;
      q.push((p+4)/5);
    }
    if(p/2==0)m[0]=min(m[0],m[p]+(p%2)*d);
    if(p/3==0)m[0]=min(m[0],m[p]+(p%3)*d);
    if(p/5==0)m[0]=min(m[0],m[p]+(p%5)*d);
    if(p<INF/d)m[0]=min(m[0],m[p]+p*d);
  }
  cout << m[0]-1 << endl;
}

return 0;
}