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 proconio::*;
use std::cmp::*;

// 割っていく方針

#[fastout]
fn run() {
    input! {
        t: usize,
        ask: [(i64, i64, i64, i64, i64); t],
    }
    for (n, a, b, c, d) in ask {
        let mut map = std::collections::BTreeMap::<i64, i64>::new();
        let mut h = std::collections::BinaryHeap::<Reverse<(i64, i64)>>::new();
        let mut ans = std::i64::MAX;
        map.insert(n, 0i64);
        h.push(Reverse((0i64, n)));
        while let Some(Reverse((x, v))) = h.pop() {
            if x > *map.get(&v).unwrap() {
                continue;
            }
            ans = std::cmp::min(ans, x.saturating_add(v.saturating_mul(d)));
            for &(cost, div) in [(a, 2i64), (b, 3i64), (c, 5i64)].iter() {
                let y = v / div;
                for &y in [y, y + 1].iter() {
                    let dis = (v - y * div).abs();
                    let x = x.saturating_add(dis.saturating_mul(d)).saturating_add(cost);
                    let po = map.entry(y).or_insert(std::i64::MAX);
                    if *po > x {
                        *po = x;
                        h.push(Reverse((x, y)));
                    }
                }
            }
        }
        println!("{}", ans);
    }
}

fn main() {
    run();
}