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)

// Code By H~$~C
#include <bits/stdc++.h>
using namespace std;

constexpr int inf = 0x3f3f3f3f;
constexpr long long lnf = 0x3f3f3f3f3f3f3f3f;

unordered_map<long long, long long> dp;
long long n, a, b, c, d;

long long dfs(long long x) {
  if (x == 0) return 0;
  if (x == 1) return d;
  if (dp.find(x) != dp.end()) return dp[x];
  long long &ans = dp[x];
  ans = lnf;
  long long A, B, C;
  A = x / 2 * 2;
  B = x / 3 * 3;
  C = x / 5 * 5;
  ans = min(ans, dfs(A / 2) + a + d * (x - A));
  ans = min(ans, dfs(B / 3) + b + d * (x - B));
  ans = min(ans, dfs(C / 5) + c + d * (x - C));
  A += 2;
  B += 3;
  C += 5;
  ans = min(ans, dfs(A / 2) + a + d * (A - x));
  ans = min(ans, dfs(B / 3) + b + d * (B - x));
  ans = min(ans, dfs(C / 5) + c + d * (C - x));
  if (x < lnf / d) ans = min(ans, x * d);
  return ans;
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr); 
  cout << fixed << setprecision(12);
  int tests;
  cin >> tests;
  while (tests--) {
    cin >> n >> a >> b >> c >> d;
    unordered_map<long long, long long> emp;
    dp = emp;
    cout << dfs(n) << endl;
  }
  return 0;
}