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)

#[allow(unused_imports)]
use proconio::{input, marker::*};

#[allow(unused_imports)]
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, HashMap, HashSet, VecDeque};

#[allow(unused_imports)]
use std::io::Write;

#[allow(unused_macros)]
macro_rules! debug {
    ($($a:expr),*) => {
        #[cfg(debug_assertions)]
        writeln!(&mut std::io::stderr(), concat!("[DEBUG] ", $(stringify!($a), "={:?} "),*), $($a),*).unwrap();
    }
}

fn sat_pow(b: usize, mut n: usize) -> usize {
    let mut res = 1usize;
    let mut mul = b;
    while n > 0 {
        if n & 1 == 1 {
            res = res.saturating_mul(mul);
        }
        mul = mul.saturating_mul(mul);
        n >>= 1;
    }
    res
}

macro_rules ! min { ( $ x : expr ) => ( $ x ) ; ( $ x : expr , $ ( $ z : expr ) ,+ ) => { { std :: cmp :: min ( $ x , min ! ( $ ( $ z ) ,* ) ) } } }

fn main() {
    use core::cmp::Reverse;
    input! {
        t: usize,
    }
    for _ in 0..t {
        input! {
            n: usize,
            a: usize,
            b: usize,
            c: usize,
            d: usize,
        }

        let mut heap = BinaryHeap::new();
        heap.push((Reverse(0), n));
        let mut dist = BTreeMap::new();
        dist.insert(n, 0);
        let neighbors = vec![(2, a), (3, b), (5, c)];
        while let Some((Reverse(cost), val)) = heap.pop() {
            if val == 1 {
                println!("{}", cost + d);
                break;
            }
            for &(base, bcost) in &neighbors {
                if val % base == 0 {
                    let kd = dist.get(&(val / base)).cloned().unwrap_or(std::usize::MAX);
                    let nd = cost.saturating_add(min!(bcost, (val - val / base).saturating_mul(d)));
                    if kd > nd {
                        dist.insert(val / base, nd);
                        heap.push((Reverse(nd), val / base));
                    }
                } else {
                    let v1 = val / base * base;
                    let v2 = v1 + base;
                    let kd = dist.get(&(v1 / base)).cloned().unwrap_or(std::usize::MAX);
                    let nd = cost.saturating_add(
                        min!(bcost, (v1 - v1 / base).saturating_mul(d)) + (val - v1) * d,
                    );
                    if kd > nd {
                        dist.insert(v1 / base, nd);
                        heap.push((Reverse(nd), v1 / base));
                    }
                    let kd = dist.get(&(v2 / base)).cloned().unwrap_or(std::usize::MAX);
                    let nd = cost.saturating_add(
                        min!(bcost, (v2 - v2 / base).saturating_mul(d)) + (v2 - val) * d,
                    );
                    if kd > nd {
                        dist.insert(v2 / base, nd);
                        heap.push((Reverse(nd), v2 / base));
                    }
                }
            }
        }
    }
}