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)

macro_rules !read_value {($iter :expr ,($($t :tt ) ,*) ) =>{($(read_value !($iter ,$t ) ) ,*) } ;($iter :expr ,[$t :tt ;$len :expr ] ) =>{(0 ..$len ) .map (|_ |read_value !($iter ,$t ) ) .collect ::<Vec <_ >>() } ;($iter :expr ,{chars :$base :expr } ) =>{read_value !($iter ,String ) .chars () .map (|c |(c as u8 -$base as u8 ) as usize ) .collect ::<Vec <usize >>() } ;($iter :expr ,{char :$base :expr } ) =>{read_value !($iter ,{chars :$base } ) [0 ] } ;($iter :expr ,chars ) =>{read_value !($iter ,String ) .chars () .collect ::<Vec <char >>() } ;($iter :expr ,char ) =>{read_value !($iter ,chars ) [0 ] } ;($iter :expr ,usize1 ) =>{read_value !($iter ,usize ) -1 } ;($iter :expr ,$t :ty ) =>{$iter .next () .unwrap () .parse ::<$t >() .unwrap () } ;}
macro_rules !input_inner {($iter :expr ) =>{} ;($iter :expr ,) =>{} ;($iter :expr ,mut $var :ident :$t :tt $($r :tt ) *) =>{let mut $var =read_value !($iter ,$t ) ;input_inner !{$iter $($r ) *} } ;($iter :expr ,mut $var :ident $($r :tt ) *) =>{input_inner !{$iter ,mut $var :usize $($r ) *} } ;($iter :expr ,$var :ident :$t :tt $($r :tt ) *) =>{let $var =read_value !($iter ,$t ) ;input_inner !{$iter $($r ) *} } ;($iter :expr ,$var :ident $($r :tt ) *) =>{input_inner !{$iter ,$var :usize $($r ) *} } ;}
macro_rules !input {(source =$s :expr ,$($r :tt ) *) =>{let mut iter =$s .split_whitespace () ;input_inner !{iter ,$($r ) *} } ;(iter =$iter :ident ,$($r :tt ) *) =>{let s ={use std ::io ::Read ;let mut s =String ::new () ;std ::io ::stdin () .read_to_string (&mut s ) .unwrap () ;s } ;let mut $iter =s .split_whitespace () ;input_inner !{$iter ,$($r ) *} } ;($($r :tt ) *) =>{let s ={use std ::io ::Read ;let mut s =String ::new () ;std ::io ::stdin () .read_to_string (&mut s ) .unwrap () ;s } ;let mut iter =s .split_whitespace () ;input_inner !{iter ,$($r ) *} } ;}
use std::collections::HashMap;
fn abssub(x: usize, y: usize) -> usize {
    if x > y {
        x - y
    } else {
        y - x
    }
}
fn main() {
    input!(t, q: [(usize, usize, usize, usize, usize); t]);
    use std::cmp::Reverse;
    for &(n, a, b, c, d) in &q {
        let mut ans = std::usize::MAX;
        let mut memo = HashMap::new();
        memo.insert(n, 0usize);
        let mut heap = std::collections::BinaryHeap::new();
        heap.push(Reverse((0usize, n)));
        while let Some(Reverse((x, m))) = heap.pop() {
            if memo.get(&m).map(|&z| z < x).unwrap_or(false) {
                continue;
            }
            ans = std::cmp::min(ans, x.saturating_add(d.saturating_mul(m)));
            for &(k, e) in &[(2, a), (3, b), (5, c)] {
                for &l in &[m / k, m / k + 1] {
                    let y = x.saturating_add(abssub(m, l * k).saturating_mul(d).saturating_add(e));
                    let z = memo.entry(l).or_insert(std::usize::MAX);
                    if *z > y {
                        *z = y;
                        heap.push(Reverse((y, l)));
                    }
                }
            }
        }
        println!("{}", ans);
    }
}