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;

#ifdef LOCAL
#include "dump.hpp"
#else
#define dump(...)
#define dumpv(...)
#endif

#define rep(i, n) for (int i = 0; i < (n); i++)
#define mins(x, y) (x = min(x, y))
#define maxs(x, y) (x = max(x, y))
using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
using vvi = vector<vi>;
using vvl = vector<vl>;
using P = pair<int, int>;
const int MOD = 1e9 + 7;
const int INF = 1001001001;
const ll LINF = 1001002003004005006ll;

ll a, b, c, d;
map<ll, ll> mp;
ll f(ll n, ll co) {
  // dump(n, co);
  if (n < 0) return LINF;
  if (n == 0) return co;
  if (n == 1) return co + d;
  if (mp[n] > 0) return co + mp[n];
  ll add = LINF;
  if (LINF / d > n) add = n * d;
  if (n % 2 == 0) {
    mins(add, f(n / 2, co) + a);
  } else {
    mins(add, f((n - 1) / 2, co) + a + d);
    mins(add, f((n + 1) / 2, co) + a + d);
  }
  if (n % 3 == 0) {
    mins(add, f(n / 3, co) + b);
  } else if (n % 3 == 1) {
    mins(add, f((n - 1) / 3, co) + b + d);
    mins(add, f((n + 2) / 3, co) + b + d * 2);
  } else {
    mins(add, f((n - 2) / 3, co) + b + d * 2);
    mins(add, f((n + 1) / 3, co) + b + d);
  }
  if (n % 5 == 0) {
    mins(add, f(n / 5, co) + c);
  } else if (n % 5 == 1) {
    mins(add, f((n - 1) / 5, co) + c + d);
    mins(add, f((n + 4) / 5, co) + c + d * 4);
  } else if (n % 5 == 2) {
    mins(add, f((n - 2) / 5, co) + c + d * 2);
    mins(add, f((n + 3) / 5, co) + c + d * 3);
  } else if (n % 5 == 3) {
    mins(add, f((n - 3) / 5, co) + c + d * 3);
    mins(add, f((n + 2) / 5, co) + c + d * 2);
  } else {
    mins(add, f((n - 4) / 5, co) + c + d * 4);
    mins(add, f((n + 1) / 5, co) + c + d);
  }
  dump(n, add, co);
  mp[n] = add - co;
  return add;
}

void solve() {
  int t;
  cin >> t;
  rep(i, t) {
    ll n;
    cin >> n >> a >> b >> c >> d;
    ll ans = f(n, 0);
    cout << ans << endl;
    mp.clear();
  }
}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(15);
  // freopen("temp.1", "r", stdin);
  solve();
  return 0;
}