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, marker::*};
use std::cmp::*;
use std::collections::*;
use superslice::*;

fn main() {
    input! {
        t: usize,
        cases: [(u64, u64, u64, u64, u64);t],
    }

    for (n, a, b, c, d) in cases {
        let mut memo: HashMap<u64, u64> = HashMap::new();
        let ans = solve(n, a, b, c, d, &mut memo);
        println!("{}", ans);
    }
}

fn solve(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(&n) = memo.get(&n) {
        return n;
    }

    let mut res = d.saturating_mul(n);

    //A
    if n % 2 == 0 {
        res = min(res, solve(n / 2, a, b, c, d, memo) + a);
    } else {
        let plus = 2 - (n % 2);
        res = min(res, solve(n / 2 + 1, a, b, c, d, memo) + a + d * plus);

        let minus = n % 2;
        res = min(res, solve(n / 2, a, b, c, d, memo) + a + d * minus);
    }

    //B
    if n % 3 == 0 {
        res = min(res, solve(n / 3, a, b, c, d, memo) + b);
    } else {
        let plus = 3 - (n % 3);
        res = min(res, solve(n / 3 + 1, a, b, c, d, memo) + b + d * plus);

        let minus = n % 3;
        res = min(res, solve(n / 3, a, b, c, d, memo) + b + d * minus);
    }

    //C
    if n % 5 == 0 {
        res = min(res, solve(n / 5, a, b, c, d, memo) + c);
    } else {
        let plus = 5 - (n % 5);
        res = min(res, solve(n / 5 + 1, a, b, c, d, memo) + c + d * plus);

        let minus = n % 5;
        res = min(res, solve(n / 5, a, b, c, d, memo) + c + d * minus);
    }

    memo.insert(n, res);
    res
}