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 sys,os,io
import math,bisect,operator
inf,mod = float('inf'),10**9+7
# sys.setrecursionlimit(10 ** 6)
from itertools import groupby,accumulate
from heapq import heapify,heappop,heappush
from collections import deque,Counter,defaultdict
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
Neo = lambda : list(map(int,input().split()))
test, = Neo()
for _ in range(test):
    n,m = Neo()
    A = Neo()
    B = Neo()
    if m > sum(A):
        print(-1)
        continue
    C,D = [],[]
    for i,j in zip(A,B):
        if j == 1:
            C += [i]
        else:
            D += [i]
    Ans = 0    
    C.sort(reverse=True)
    D.sort(reverse=True)
    # print(C,D)
    E,F = [],[]
    k = 0
    while C and D and m > 0:
        if len(C) > 1:
            if m > C[0]:
                if C[0]+C[1] > D[0]:
                    m -= C[0]+C[1]
                    E += [C[0]]
                    E += [C[1]]
                    C.pop(0)
                    C.pop(0)
                    Ans += 2
                else:
                    m -= D[0]
                    F += [D[0]]
                    D.pop(0)
                    Ans += 2
            else:
                m -= C[0]
                E += [C[0]]
                C.pop(0)
                Ans += 1
        else:
            t = 0
            while m > 0 and D:
                if m < C[0]:
                    Ans += 1
                    m -= C[0]
                    E += [C[0]]
                    C.pop(0)
                    break
                t = 1    
                Ans += 2
                m -= D[0]
                F += [D[0]]
                D.pop(0)
            if C:
                Ans += 1
                m -= C.pop(0)
            if -m in C and t:
                Ans -= 1
            k = 1  
    if k:
        print(Ans)
        continue
    while m > 0 and C:
        m -= C[0]
        E += [C[0]]
        Ans += 1
        C.pop(0)    
    while m > 0 and D:
        m -= D[0]
        F += [D[0]]
        Ans += 2
        D.pop(0) 
    # print(E,F,m)    
    if -m in F:
        Ans -= 2
    elif -m in E:
        Ans -= 1
    print(Ans)  
                