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::collections::{BinaryHeap,BTreeMap};
use std::cmp::Reverse;
fn main() {
    proconio::input! {
        t: usize,
        nabcd: [(i64,i64,i64,i64,i64);t],
    }
    for (n,a,b,c,d) in nabcd {
        let mut e = BTreeMap::new();
        let mut q = BinaryHeap::new();
        q.push((Reverse(0),n));
        while let Some((Reverse(s),i))=q.pop() {
            if e.contains_key(&i) {continue;}
            e.insert(i, s);
            if i==0 {
                break;
            }
            for (k,v) in &[(2,a),(3,b),(5,c)] {
                let r=i%k;
                let c=r*d+v;
                let j=i/k;
                if c>=d.saturating_mul(i-j) {
                    q.push((Reverse(s+i*d),0));
                }
                else {
                    if !e.contains_key(&j) {
                        q.push((Reverse(s+c),j));
                    }
                    if r>0 {
                        let c=(k-r)*d+v;
                        let j=j+1;
                        if c<d.saturating_mul(i-j) && !e.contains_key(&j) {
                            q.push((Reverse(s+c),j));
                        }
                    }
                }
            }
        }
        println!("{}",e.get(&0).unwrap_or(&-1));
    }
}