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 collections import defaultdict
import heapq

def a(N,A,B,C,D):
    E = defaultdict(lambda:N*D)
    arr = [(2,A),(3,B),(5,C)]
    # Nから0までのコストを計算する
    # ヒープは最小の要素が先頭に来る
    # D[0]を超えたら打ち切る
    aa = 0
    bb = 0
    num = 0
    H = [N]
    E[N] = 0
    while len(H) != 0:
        num = heapq.heappop(H)
        if E[num] >= E[0]:
            continue
        E[0] = min(E[0],E[num] + D * num)
        for k,v in arr:
            aa = num // k
            bb = E[num] + D * (num - aa * k) + v
            if E[aa] > bb:
                E[aa] = bb
                heapq.heappush(H,aa)
            aa += 1
            bb = E[num] + D * (aa * k - num) + v
            if E[aa] > bb:
                E[aa] = bb
                heapq.heappush(H,aa)
    print(E[0])

T = int(input())
A = [0]*T
for i in range(T):
    A[i] = tuple(map(int,input().split()))
    
for i in range(T):
    a(*A[i])