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 PyPy3 (7.3.0) (judge verdict: Accepted)

check=dict()
N,z,d=0,[],0
#ttfは235、chaはabc

def dfs_sub(n,ttf,cha,d):
    global check

    #-1ずつしていく場合
    check[0]=min(check[n]+n*d,check[0])

    #最寄りまで-してく場合
    x1,x2=n//ttf,n%ttf
    if x1 in check:
        if check[x1]>check[n]+x2*d+cha:
            check[x1]=check[n]+x2*d+cha
            dfs(x1)
    else:
        check[x1]=check[n]+x2*d+cha
        dfs(x1)

    #最寄りまで+1してく場合
    x1,x2=n//ttf+1,ttf-n%ttf
    if x1 in check:
        if check[x1]>check[n]+x2*d+cha:
            check[x1]=check[n]+x2*d+cha
            dfs(x1)
    else:
        check[x1]=check[n]+x2*d+cha
        dfs(x1)


def dfs(n):
    global b,d
    for i,j in zip([2,3,5],b):
        dfs_sub(n,i,j,d)

t=int(input())
for i in range(t):
    N,*b,d=map(int,input().split())
    z=list(zip([2,3,5],b))
    check={N:0,0:N*d}
    dfs(N)
    print(check[0])