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: Time Limit Exceeded)

#![allow(unused_imports)]
use itertools::Itertools;
use proconio::{input, marker::*};
use std::cmp::*;
use std::collections::*;
use superslice::*;

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

    for (n, a, b, c, d) in cases {
        solve(n, a, b, c, d);
    }
}

fn solve(n: u64, a: u64, b: u64, c: u64, d: u64) {
    let mut dp: HashMap<u64, u64> = HashMap::new();

    let mut q: VecDeque<(u64, u64)> = VecDeque::new();
    q.push_back((n, 0));

    while let Some((k, v)) = q.pop_front() {
        let &bef = dp.get(&k).unwrap_or(&std::u64::MAX);
        if bef <= v {
            continue;
        }
        dp.insert(k, v);

        if k == 0 {
            continue;
        }

        //A
        if k % 2 == 0 {
            if a / d < k / 2 {
                q.push_back((k / 2, v + a));
            } else {
                q.push_back((k / 2, v + k / 2 * d));
            }
        } else if k > 2 {
            let plus = 2 - (k % 2);
            let plus_diff = (k + plus) / 2;
            if a / d < plus_diff {
                q.push_back((k + plus - plus_diff, v + d * plus + a));
            } else {
                q.push_back((k + plus - plus_diff, v + d * plus + d * plus_diff));
            }

            let minus = k % 2;
            let minus_diff = (k - minus) / 2;
            if a / d < minus_diff {
                q.push_back((k - minus - minus_diff, v + d * minus + a));
            } else {
                q.push_back((k - minus - minus_diff, v + d * minus + d * minus_diff));
            }
        } else {
            q.push_back((k - k, v + k * d));
        }

        //B
        if k % 3 == 0 {
            if b / d < k * 2 / 3 {
                q.push_back((k / 3, v + b));
            } else {
                q.push_back((k / 3, v + k * 2 / 3 * d));
            }
        } else if k > 3 {
            let plus = 3 - (k % 3);
            let plus_diff = (k + plus) / 3 * 2;
            if b / d < plus_diff {
                q.push_back((k + plus - plus_diff, v + d * plus + b));
            } else {
                q.push_back((k + plus - plus_diff, v + d * plus + d * plus_diff));
            }

            let minus = k % 3;
            let minus_diff = (k - minus) / 3 * 2;
            if b / d < minus_diff {
                q.push_back((k - minus - minus_diff, v + d * minus + b));
            } else {
                q.push_back((k - minus - minus_diff, v + d * minus + d * minus_diff));
            }
        } else {
            q.push_back((k - k, v + k * d));
        }

        //C
        if k % 5 == 0 {
            if c / d < k * 4 / 5 {
                q.push_back((k / 5, v + c));
            } else {
                q.push_back((k / 5, v + k * 4 / 5 * d));
            }
        } else if k > 5 {
            let plus = 5 - (k % 5);
            let plus_diff = (k + plus) / 5 * 4;
            if c / d < plus_diff {
                q.push_back((k + plus - plus_diff, v + d * plus + c));
            } else {
                q.push_back((k + plus - plus_diff, v + d * plus + d * plus_diff));
            }

            let minus = k % 5;
            let minus_diff = (k - minus) / 5 * 4;
            if c / d < minus_diff {
                q.push_back((k - minus - minus_diff, v + d * minus + c));
            } else {
                q.push_back((k - minus - minus_diff, v + d * minus + d * minus_diff));
            }
        } else {
            q.push_back((k - k, v + k * d));
        }
    }

    println!("{}", dp.get(&0).unwrap());
}