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 Rust (1.42.0) (judge verdict: Accepted)

use std::cmp::Reverse;
use std::collections::{BinaryHeap, HashMap};

use proconio::input;

fn solve(target: i64, a: u128, b: u128, c: u128, d: u128) {
    let mut heap = BinaryHeap::new();
    let mut best = HashMap::new();
    heap.push((Reverse(0), target));
    best.insert(target, 0);

    fn push(
        heap: &mut BinaryHeap<(Reverse<u128>, i64)>,
        best: &mut HashMap<i64, u128>,
        m: i64,
        cost: u128,
    ) {
        let now_best = best.entry(m).or_insert(std::u128::MAX);
        if cost < *now_best {
            *now_best = cost;
            heap.push((Reverse(cost), m));
        }
    }

    while let Some((Reverse(cost), n)) = heap.pop() {
        if n == 0 {
            println!("{}", cost);
            return;
        }
        if n < 0 {
            continue;
        }
        if n % 2 == 0 {
            let m = n / 2;
            let cost = cost + a.min(m as u128 * d);
            push(&mut heap, &mut best, m, cost);
        } else {
            push(&mut heap, &mut best, n - 1, cost + d);
            push(&mut heap, &mut best, n + 1, cost + d);
        }
        if n % 3 == 0 {
            let m = n / 3;
            let cost = cost + b.min((m * 2) as u128 * d);
            push(&mut heap, &mut best, m, cost);
        } else if n % 3 == 1 {
            push(&mut heap, &mut best, n - 1, cost + d);
            push(&mut heap, &mut best, n + 2, cost + d * 2);
        } else {
            push(&mut heap, &mut best, n - 2, cost + d * 2);
            push(&mut heap, &mut best, n + 1, cost + d);
        }
        if n % 5 == 0 {
            let m = n / 5;
            let cost = cost + c.min((m * 4) as u128 * d);
            push(&mut heap, &mut best, m, cost);
        } else if n % 5 == 1 {
            push(&mut heap, &mut best, n - 1, cost + d);
            push(&mut heap, &mut best, n + 4, cost + d * 4);
        } else if n % 5 == 2 {
            push(&mut heap, &mut best, n - 2, cost + d * 2);
            push(&mut heap, &mut best, n + 3, cost + d * 3);
        } else if n % 5 == 3 {
            push(&mut heap, &mut best, n - 3, cost + d * 3);
            push(&mut heap, &mut best, n + 2, cost + d * 2);
        } else {
            push(&mut heap, &mut best, n - 4, cost + d * 4);
            push(&mut heap, &mut best, n + 1, cost + d);
        }
    }
}

fn main() {
    input! {
        t: u64,
        cases: [(i64, u128, u128, u128, u128); t],
    };

    for case in cases {
        solve(case.0, case.1, case.2, case.3, case.4);
    }
}