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;

const long long INF = 10000000000000000;
long long n;
int a, b, c, d;
map<long long, long long> memo;

long long rec(long long n) {
    if (n == 0) return 0;
    if (memo.count(n)) return memo[n];
    long long res = n * d;
    if (n % 5 == 0) {
        res = min(res, rec(n / 5) + c);
    } else if (n % 5 == 1) {
        res = min(res, rec(n / 5) + c + d);
    } else if (n % 5 == 2) {
        res = min(res, rec(n / 5) + c + d + d);
    } else if (n % 5 == 3) {
        res = min(res, rec((n / 5) + 1) + c + d + d);
    } else if (n % 5 == 4) {
        res = min(res, rec((n / 5) + 1) + c + d);
    }
    if (n % 3 == 0) {
        res = min(res, rec(n / 3) + b);
    } else if (n % 3 == 1) {
        res = min(res, rec(n / 3) + b + d);
    } else if (n % 3 == 2) {
        res = min(res, rec((n / 3) + 1) + b + d);
    }
    if (n % 2 == 0) {
        res = min(res, rec(n / 2) + a);
    } else if (n % 2 == 1) {
        res = min(res, rec(n / 2) + a + d);
        res = min(res, rec((n / 2) + 1) + a + d);
    }
    memo[n] = res;
    return res;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    for (int i = 0; i < t; i++) {
        cin >> n;
        cin >> a >> b >> c >> d;
        memo.clear();
        memo[0] = 0;
        memo[1] = d;
        cout << rec(n) << '\n';
    }
    return 0;
}