Helmets in Night Light
Pak Chanek is the chief of a village named Khuntien. On one night filled with lights, Pak Chanek has a sudden and important announcement that needs to be notified to all of the $$$n$$$ residents in Khuntien.First, Pak Chanek shares the announcement directly to one or more residents with a cost of $$$p$$$ for each person. After that, the residents can share the announcement to other residents using a magical helmet-shaped device. However, there is a cost for using the helmet-shaped device. For each $$$i$$$, if the $$$i$$$-th resident has got the announcement at least once (either directly from Pak Chanek or from another resident), he/she can share the announcement to at most $$$a_i$$$ other residents with a cost of $$$b_i$$$ for each share.If Pak Chanek can also control how the residents share the announcement to other residents, what is the minimum cost for Pak Chanek to notify all $$$n$$$ residents of Khuntien about the announcement?
In the first test case, the following is a possible optimal strategy:   Pak Chanek shares the announcement directly to the $$$3$$$-rd, $$$5$$$-th, and $$$6$$$-th resident. This requires a cost of $$$p+p+p=3+3+3=9$$$.  The $$$3$$$-rd resident shares the announcement to the $$$1$$$-st and $$$2$$$-nd resident. This requires a cost of $$$b_3+b_3=2+2=4$$$.  The $$$2$$$-nd resident shares the announcement to the $$$4$$$-th resident. This requires a cost of $$$b_2=3$$$. The total cost is $$$9+4+3=16$$$. It can be shown that there is no other strategy with a smaller cost.
Each test contains multiple test cases. The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The following lines contain the description of each test case.The first line contains two integers $$$n$$$ and $$$p$$$ ($$$1 \leq n \leq 10^5$$$; $$$1 \leq p \leq 10^5$$$) — the number of residents and the cost for Pak Chanek to share the announcement directly to one resident.The second line contains $$$n$$$ integers $$$a_1,a_2,a_3,\ldots,a_n$$$ ($$$1\leq a_i\leq10^5$$$) — the maximum number of residents that each resident can share the announcement to.The third line contains $$$n$$$ integers $$$b_1,b_2,b_3,\ldots,b_n$$$ ($$$1\leq b_i\leq10^5$$$) — the cost for each resident to share the announcement to one other resident.It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
For each test case, output a line containing an integer representing the minimum cost to notify all $$$n$$$ residents of Khuntien about the announcement.
import sys
input = sys.stdin.readline

def readList():
    return list(map(int, input().split()))

def readInt():
    return int(input())

def readInts():
    return map(int, input().split())

def readStr():
    return input().strip()

def solve():
    (n, p) = readInts()
    (a, b) = (readList(), readList())
    idx = sorted([i for i in range(n)], key=lambda x: b[x])
    while idx and b[idx[-1]] >= p:
        idx.pop()
    (cost, curr) = (p, n - 1)
    for i in idx:
        if not curr:
            break
        cnt = min(curr, a[i])
        cost += cnt * b[i]
        curr -= cnt
    return cost + curr * p
for _ in range(int(input())):
    print(solve())
t = int(input())
while t > 0:
    t -= 1
    (n, p) = map(int, input().split())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))
    order = [i for i in range(n)]
    order.sort(key=lambda x: b[x])
    res = p
    n -= 1
    pos = 0
    while n > 0 and b[order[pos]] < p:
        if n <= a[order[pos]]:
            res += b[order[pos]] * n
            n = 0
        else:
            n -= a[order[pos]]
            res += a[order[pos]] * b[order[pos]]
        pos += 1
    if n != 0:
        res += n * p
        n = 0
    print(res)
def read_int():
    return int(input())

def read_list():
    return list(map(int, input().split()))

def main():
    t = int(input())
    while t > 0:
        solve()
        t -= 1

def solve():
    (n, p) = read_list()
    a = read_list()
    b = read_list()
    c = list(zip(b, a))
    c.sort()
    pay = p
    cur = 0
    n -= 1
    while n > 0 and p > c[cur][0]:
        if c[cur][1] >= n:
            pay += n * c[cur][0]
            n -= n
        else:
            pay += c[cur][1] * c[cur][0]
            n -= c[cur][1]
        cur += 1
    if n > 0:
        pay += p * n
    print(pay)
if __name__ == '__main__':
    main()