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)

use std::cmp::*;
use std::collections::HashMap;
use std::io::*;

fn dfs(dp: &mut HashMap<i64, i64>, x: i64, a: i64, b: i64, c: i64, d: i64) -> i64 {
    if dp.contains_key(&x) {
        dp[&x]
    } else {
        let mut res = x.saturating_mul(d);
        res = min(res, dfs(dp, x / 2, a, b, c, d) + a + x % 2 * d);
        if x % 2 != 0 {
            res = min(res, dfs(dp, x / 2 + 1, a, b, c, d) + a + d);
        }
        res = min(res, dfs(dp, x / 3, a, b, c, d) + b + x % 3 * d);
        if x % 3 != 0 {
            res = min(res, dfs(dp, x / 3 + 1, a, b, c, d) + b + (3 - x % 3) * d);
        }
        res = min(res, dfs(dp, x / 5, a, b, c, d) + c + x % 5 * d);
        if x % 5 != 0 {
            res = min(res, dfs(dp, x / 5 + 1, a, b, c, d) + c + (5 - x % 5) * d);
        }
        dp.insert(x, res);
        res
    }
}

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let t: usize = itr.next().unwrap().parse().unwrap();
    for _ in 0..t {
        let n: i64 = itr.next().unwrap().parse().unwrap();
        let a: i64 = itr.next().unwrap().parse().unwrap();
        let b: i64 = itr.next().unwrap().parse().unwrap();
        let c: i64 = itr.next().unwrap().parse().unwrap();
        let d: i64 = itr.next().unwrap().parse().unwrap();
        let mut dp = HashMap::<i64, i64>::new();
        dp.insert(0, 0);
        dp.insert(1, d);
        println!("{}", dfs(&mut dp, n, a, b, c, d));
    }
}