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 itertools::Itertools;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;

struct Solver {
    memo: std::collections::BTreeMap<u64, u64>,
    a: u64,
    b: u64,
    c: u64,
    d: u64,
}
impl Solver {
    fn new(a: u64, b: u64, c: u64, d: u64) -> Self {
        Self {
            memo: std::collections::BTreeMap::new(),
            a: a,
            b: b,
            c: c,
            d: d,
        }
    }
    fn solve(&mut self, x: u64) -> u64 {
        if x == 0 {
            0
        } else if x == 1 {
            self.d
        } else if let Some(y) = self.memo.get(&x) {
            *y
        } else {
            let mut res = x.saturating_mul(self.d);
            use std::cmp::min;
            if x % 2 == 0 {
                res = min(res, self.solve(x / 2) + self.a);
            } else {
                let y = (x / 2) * 2;
                res = min(res, self.solve(y / 2) + self.a + (x - y) * self.d);
                res = min(res, self.solve(y / 2 + 1) + self.a + (y + 2 - x) * self.d);
            }
            if x % 3 == 0 {
                res = min(res, self.solve(x / 3) + self.b);
            } else {
                let y = (x / 3) * 3;
                res = min(res, self.solve(y / 3) + self.b + (x - y) * self.d);
                res = min(res, self.solve(y / 3 + 1) + self.b + (y + 3 - x) * self.d);
            }
            if x % 5 == 0 {
                res = min(res, self.solve(x / 5) + self.c);
            } else {
                let y = (x / 5) * 5;
                res = min(res, self.solve(y / 5) + self.c + (x - y) * self.d);
                res = min(res, self.solve(y / 5 + 1) + self.c + (y + 5 - x) * self.d);
            }
            self.memo.insert(x, res);
            res
        }
    }
}

fn main() {
    input! {
        t: usize,
    }
    for _ in 0..t {
        input! { n: u64, a: u64, b :u64, c: u64, d: u64 }
        let mut solver = Solver::new(a, b, c, d);
        println!("{}", solver.solve(n));
    }
}