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 std::io;
use std::cmp;
use std::collections::BinaryHeap;
use std::collections::HashMap;
use std::collections::HashSet;
use std::f64::consts::PI;
use std::str::FromStr;
use std::cmp::Reverse;

#[allow(dead_code)]
type Error = Box<dyn std::error::Error>;

fn read_line() -> String {
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).ok();
    buf
}

struct Scanner {
    buff: Vec<String>,
    pos: usize,
}

#[allow(dead_code)]
impl Scanner {
    pub fn new() -> Self {
        Scanner {
            buff: Vec::new(),
            pos: 0,
        }
    }

    pub fn next<T: FromStr>(&mut self) -> T {
        while self.buff.len() <= self.pos {
            self.pos = 0;
            self.buff = read_line().split_whitespace().map(|s| s.to_owned()).collect();
        }

        let result = self.buff[self.pos].parse::<T>();
        self.pos += 1;
        result.ok().unwrap()
    }

    pub fn next_line(&mut self) -> String {
        let result  = read_line().trim().to_owned();
        self.buff.clear();
        self.pos = 0;

        result
    }

    pub fn next_vec<T: FromStr>(&mut self) -> Vec<T> {
        self.buff.clear();
        self.pos = 0;
        read_line().trim().split_whitespace().map(|s| s.parse().ok().unwrap()).collect()
    }
    
    pub fn next_line_chars(&mut self) -> Vec<char> {
        let result  = read_line().trim().chars().collect();
        self.buff.clear();
        self.pos = 0;

        result
    }
}

fn func(n: u64, a: u64, b: u64, c: u64, d: u64, memo: &mut HashMap<u64, u64>) -> u64 {
    if n == 0 {
        return 0;
    }
    if n == 1 {
        return d;
    }

    if let Some(&v) = memo.get(&n) {
        return v;
    }
    let n2 = n / 2 * 2;
    let n2u = (n+1) / 2 * 2;
    let n3 = n / 3 * 3;
    let n3u = (n+2) / 3 * 3;
    let n5 = n / 5 * 5;
    let n5u = (n+4) / 5 * 5;

    let mut ans = 1_000_000_000_000_000_010;
    if n < ans /d {
        ans = n * d;
    }
    ans = cmp::min(ans, (n - n2) * d + a + func(n2/2, a, b, c, d, memo));
    ans = cmp::min(ans, (n2u - n) * d + a + func(n2u/2, a, b, c, d, memo));
    ans = cmp::min(ans, (n - n3) * d + b + func(n3/3, a, b, c, d, memo));
    ans = cmp::min(ans, (n3u - n) * d + b + func(n3u/3, a, b, c, d, memo));
    ans = cmp::min(ans, (n - n5) * d + c + func(n5/5, a, b, c, d, memo));
    ans = cmp::min(ans, (n5u - n) * d + c + func(n5u/5, a, b, c, d, memo));

    memo.insert(n, ans);

    ans
}


fn main() {
    let mut sc = Scanner::new();
    let t: usize = sc.next();
    for _i in 0..t {
        let n: u64 = sc.next();
        let a: u64 = sc.next();
        let b: u64 = sc.next();
        let c: u64 = sc.next();
        let d: u64 = sc.next();

        let mut memo = HashMap::new();
        println!("{}", func(n, a, b, c, d, &mut memo));
    }
}