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(clippy::needless_range_loop)]
#![allow(unused_macros)]
#![allow(dead_code)]
#![allow(unused_imports)]

use itertools::Itertools;
use proconio::input;
use proconio::marker::*;

struct Memoized {
    memo: std::collections::BTreeMap<usize, usize>,
}

impl Memoized {
    fn func(&mut self, x: usize, a: usize, b: usize, c: usize, d: usize) -> usize {
        if let Some(&memo) = self.memo.get(&x) {
            return memo;
        }

        if x <= 1 {
            return d * x;
        }

        let &ans = [
            x.saturating_mul(d),
            self.func(x / 2, a, b, c, d) + x % 2 * d + a,
            self.func((x + 1) / 2, a, b, c, d) + (2 - x % 2) % 2 * d + a,
            self.func(x / 3, a, b, c, d) + x % 3 * d + b,
            self.func((x + 2) / 3, a, b, c, d) + (3 - x % 3) % 3 * d + b,
            self.func(x / 5, a, b, c, d) + x % 5 * d + c,
            self.func((x + 4) / 5, a, b, c, d) + (5 - x % 5) % 5 * d + c,
        ]
        .iter()
        .min()
        .unwrap();
        self.memo.insert(x, ans);
        ans
    }

    fn new() -> Self {
        Self {
            memo: std::collections::BTreeMap::new(),
        }
    }
}

fn main() {
    input! {
        testcases: [(usize, usize, usize, usize, usize)],
    }

    for &(n, a, b, c, d) in &testcases {
        let mut memo = Memoized::new();
        println!("{}", memo.func(n, a, b, c, d));
    }
}