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 Python (3.8.2) (judge verdict: Accepted)

#! /usr/bin/env python3

import sys
sys.setrecursionlimit(10 ** 6)

tt = int(input())
for _ in range(tt):
    n, a, b, c, d = map(int, input().split())
    memo = {}
    def solve(k):
        if k == 0:
            return 0
        elif k == 1:
            return d
        elif k in memo:
            return memo[k]
        else:
            res = k * d
            r = k % 2
            for x in [k - r, k - r + 2]:
                if x >= 1 and x // 2 < k:
                    res = min(res, d * abs(k - x) + a + solve(x // 2))
            r = k % 3
            for x in [k - r, k - r + 3]:
                if x >= 1 and x // 3 < k:
                    res = min(res, d * abs(k - x) + b + solve(x // 3))
            r = k % 5
            for x in [k - r, k - r + 5]:
                if x >= 1 and x // 5 < k:
                    res = min(res, d * abs(k - x) + c + solve(x // 5))
            memo[k] = res
            return memo[k]
    print(solve(n))