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)]
#![allow(dead_code)]

use proconio::input;
use std::cmp::Reverse;
use std::collections::*;

fn main() {
    input! {
        tt:usize
    }
    for _ in 0..tt {
        input! {
            n:i64,
            a:i64,
            b:i64,
            c:i64,
            d:i64,
        }
        let divs = vec![(2, a), (3, b), (5, c)];
        let mut dp = BTreeMap::new();
        dp.insert(n, 0);
        let mut heap = BinaryHeap::new();
        heap.push(Reverse((0, n)));
        while let Some(Reverse((sum, now))) = heap.pop() {
            if sum > dp[&now] {
                continue;
            }
            if now < (1e+18 as i64) / d {
                let zero = dp.entry(0).or_insert(std::i64::MAX);
                *zero = std::cmp::min(*zero, sum + d * now);
            }
            if now > 1 {
                for &(div, cost) in divs.iter() {
                    let (next_floor, next_ceil) = (now / div, (now + div - 1) / div);
                    for &next in [next_ceil, next_floor].iter() {
                        let next_ref = dp.entry(next).or_insert(std::i64::MAX);
                        let next_cost = sum + cost + (now - next * div).abs() * d;
                        if next_cost < *next_ref {
                            *next_ref = next_cost;
                            heap.push(Reverse((next_cost, next)));
                        }
                    }
                }
            }
        }
        println!("{}", dp[&0]);
    }
}