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.

Example

Input

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


Output

20
19
26
3821859835
23441258666
from heapq import heapify, heappop, heappush
from collections import defaultdict
Inf = 10**28
inpl = lambda: list(map(int,input().split()))
T = int(input())
for _ in range(T):
    N, A, B, C, D = inpl()
    visited = defaultdict(lambda: Inf)
    ans = Inf
    pool = [(0,N)] # coin, number
    while pool:
        c, n = heappop(pool)
        if c >= ans:
            break
        elif c > visited[n]:
            continue
        q = c + D*n
        if q < ans:
            ans = q

        for mag, cost in [(2,A), (3,B), (5,C)]:
            m = (n+mag//2)//mag
            q = c + abs(n - mag*m)*D + cost
            if q < ans and q < visited[m]:
                visited[m] = q
                heappush(pool, (q, m))

    print(ans)
