Polycarp often uses his smartphone. He has already installed n applications on it. Application with number i takes up a_i units of memory.

Polycarp wants to free at least m units of memory (by removing some applications).

Of course, some applications are more important to Polycarp than others. He came up with the following scoring system — he assigned an integer b_i to each application: 

  * b_i = 1 — regular application; 
  * b_i = 2 — important application. 



According to this rating system, his phone has b_1 + b_2 + … + b_n convenience points.

Polycarp believes that if he removes applications with numbers i_1, i_2, …, i_k, then he will free a_{i_1} + a_{i_2} + … + a_{i_k} units of memory and lose b_{i_1} + b_{i_2} + … + b_{i_k} convenience points.

For example, if n=5, m=7, a=[5, 3, 2, 1, 4], b=[2, 1, 1, 2, 1], then Polycarp can uninstall the following application sets (not all options are listed below): 

  * applications with numbers 1, 4 and 5. In this case, it will free a_1+a_4+a_5=10 units of memory and lose b_1+b_4+b_5=5 convenience points; 
  * applications with numbers 1 and 3. In this case, it will free a_1+a_3=7 units of memory and lose b_1+b_3=3 convenience points. 
  * applications with numbers 2 and 5. In this case, it will free a_2+a_5=7 memory units and lose b_2+b_5=2 convenience points. 



Help Polycarp, choose a set of applications, such that if removing them will free at least m units of memory and lose the minimum number of convenience points, or indicate that such a set does not exist.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

The first line of each test case contains two integers n and m (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ m ≤ 10^9) — the number of applications on Polycarp's phone and the number of memory units to be freed.

The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the number of memory units used by applications.

The third line of each test case contains n integers b_1, b_2, …, b_n (1 ≤ b_i ≤ 2) — the convenience points of each application.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case, output on a separate line: 

  * -1, if there is no set of applications, removing which will free at least m units of memory; 
  * the minimum number of convenience points that Polycarp will lose if such a set exists. 

Example

Input


5
5 7
5 3 2 1 4
2 1 1 2 1
1 3
2
1
5 10
2 3 2 3 2
1 2 1 2 1
4 10
5 1 3 4
1 2 1 2
4 5
3 2 1 2
2 1 2 1


Output


2
-1
6
4
3

Note

In the first test case, it is optimal to remove applications with numbers 2 and 5, freeing 7 units of memory. b_2+b_5=2.

In the second test case, by removing the only application, Polycarp will be able to clear only 2 of memory units out of the 3 needed.

In the third test case, it is optimal to remove applications with numbers 1, 2, 3 and 4, freeing 10 units of memory. b_1+b_2+b_3+b_4=6.

In the fourth test case, it is optimal to remove applications with numbers 1, 3 and 4, freeing 12 units of memory. b_1+b_3+b_4=4.

In the fifth test case, it is optimal to remove applications with numbers 1 and 2, freeing 5 units of memory. b_1+b_2=3.
import io
import os

from bisect import bisect_left


def solve1(N, M, A, B):
    inf = float("inf")
    best = inf
    for mask in range(1 << N):
        total = 0
        cost = 0
        for pos in range(N):
            if (1 << pos) & mask:
                total += A[pos]
                cost += B[pos]
        if total >= M:
            best = min(best, cost)
    if best == inf:
        return -1

    return best


def solve(N, M, A, B):
    A1 = []
    A2 = []
    for a, b in zip(A, B):
        if b == 1:
            A1.append(a)
        else:
            A2.append(a)
    A1.sort(reverse=True)
    A2.sort(reverse=True)
    total1 = sum(A1)
    total2 = sum(A2)
    if total1 + total2 < M:
        return -1

    pref1 = [0]
    for x in A1:
        pref1.append(pref1[-1] + x)
    pref2 = [0]
    for x in A2:
        pref2.append(pref2[-1] + x)
    best = float("inf")
    for i in range(len(A1) + 1):
        if pref1[i] >= M:
            best = min(best, i)
    for i in range(len(A2) + 1):
        if pref2[i] >= M:
            best = min(best, 2 * i)

    for i in range(len(A1) + 1):
        target = M - pref1[i]
        if target >= 0:
            index = bisect_left(pref2, target)
            if index < len(pref2):
                assert pref1[i] + pref2[index] >= M

                cost = i + 2 * index
                best = min(best, cost)

    return best


DEBUG = False
if DEBUG:
    import random

    random.seed(0)
    for _ in range(1000):
        N = random.randint(1, 10)
        M = random.randint(1, 10)
        A = [random.randint(1, 10) for i in range(N)]
        B = [random.randint(1, 2) for i in range(N)]
        print("tc", _, N, A, B)
        ans1 = solve1(N, M, A, B)
        ans = solve(N, M, A, B)
        assert ans1 == ans
        print(ans1)
    exit()


if __name__ == "__main__":
    input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

    TC = int(input())
    for tc in range(1, TC + 1):
        N, M = [int(x) for x in input().split()]
        A = [int(x) for x in input().split()]
        B = [int(x) for x in input().split()]
        ans = solve(N, M, A, B)
        print(ans)
