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;

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

const long long INF = 1LL << 60;

int main() {
  int T;
  cin >> T;
  
  long long N[T], A[T], B[T], C[T], D[T];
  
  for (int i = 0; i < T; i++) {
    cin >> N[i] >> A[i] >> B[i] >> C[i] >> D[i];

    long long coinmin = INF;

    long long coin0 = D[i];
    long long coinn1 = D[i] + D[i];
    long long coinp1 = D[i] + D[i];
    long long N0 = N[i];
    long long Nn1 = N[i] - 1;
    long long Np1 = N[i] + 1;    
    // N
    while (N0 % 2 == 0) {
      N0 = N0 / 2;
      coin0 += A[i];
    }
    while (N0 % 3 == 0) {
      N0 = N0 / 3;
      coin0 += B[i];
    }
    while (N0 % 5 == 0) {
      N0 = N0 / 5;
      coin0 += C[i];
    }
    if (N0 == 1) coinmin = coin0;
    // N - 1
    while (Nn1 % 2 == 0) {
      Nn1 = Nn1 / 2;
      coinn1 += A[i];
    }
    while (Nn1 % 3 == 0) {
      Nn1 = Nn1 / 3;
      coinn1 += B[i];
    }
    while (Nn1 % 5 == 0) {
      Nn1 = Nn1 / 5;
      coinn1 += C[i];
    }
    if (Nn1 == 1) chmin(coinmin, coinn1);    
    // N + 1
    while (Np1 % 2 == 0) {
      Np1 = Np1 / 2;
      coinp1 += A[i];
    }
    while (Np1 % 3 == 0) {
      Np1 = Np1 / 3;
      coinp1 += B[i];
    }
    while (Np1 % 5 == 0) {
      Np1 = Np1 / 5;
      coinp1 += C[i];
    }
    if (Np1 == 1) chmin(coinmin, coinp1);
    
    cout << coinmin << endl;
  }
}