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 __int128 bll;
typedef long long lol;
typedef pair<bll,bool> pbb;
int _w;

lol A , B , C , D , N;
bll ans;

map<pbb,bll> mp;

void print( bll x ) {
  int S[64] , tp;
  tp = 0;
  while( x ) 
    S[++tp] = x % 10 , x /= 10;
  while( tp ) putchar( S[tp--] + '0' );
}

void dfs( bll now , bll cost , bool flag ) {
  if( cost >= ans ) return;
  ans = min( ans , cost + now * D );
  if( !mp.count( pbb( now , flag ) ) || mp[pbb( now , flag )] > cost ) mp[pbb( now , flag )] = cost;
  else return;
  if( now % 5 == 0 ) dfs( now / 5 , cost + C , 1 );
  if( now % 3 == 0 ) dfs( now / 3 , cost + B , 1 );
  if( !(now&1) ) dfs( now / 2 , cost + A , 1 );
  if( flag ) {
    dfs( now + 1 , cost + D , 0 );
    if( now > 0 ) dfs( now - 1 , cost + D , 0 );
    dfs( now + 2 , cost + 2 * D , 0 );
    if( now > 1 ) dfs( now - 2 , cost + 2 * D , 0 );
  }
}


int main( void ) {
  int T;
  _w = scanf("%d",&T);
  while( T-- ) {
    _w = scanf("%lld%lld%lld%lld%lld",&N,&A,&B,&C,&D);
    mp.clear();
    ans = (bll)D * N;
    dfs( N , 0 , 1 );
    print( ans );
    putchar('\n');
  }
  return 0;
}